1. 相同的树

原题链接

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        }
        if ((p != null && q == null) || (p == null && q != null)) {
            return false;
        }
        if (p.val != q.val) {
            return false;
        }
        return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);
    }
}

2. 二叉树的最近公共祖先

原题链接

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    public TreeNode ans;

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        ans = null;
        dfs(root,p,q);
        return ans;
    }

    public boolean dfs(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return false;
        }
        boolean l = dfs(root.left,p,q);
        boolean r = dfs(root.right,p,q);
        if (l == true && r == true && ans == null) {
            ans = root;
            return true;
        }
        if ((l == true || r == true) && (root == p || root == q) && ans == null) {
            ans = root;
            return true;
        }
        if (root == p || root == q) {
            return true;
        }
        return l || r;
    }

}

3. 二叉搜索树中第K小的元素

通过双端队列duque 中序遍历

二叉搜索树中第K小的元素

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            --k;
            if (k == 0) {
                break;
            }
            root = root.right;
        }
        return root.val;
    }
}

4. 二叉树的锯齿形层序遍历

原题链接

new LinkedList(levelList)双端队列复制 数组

需要左右顺序，考虑双端队列

class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> ans = new LinkedList<List<Integer>>();
        if (root == null) {
            return ans;
        }

        Queue<TreeNode> nodeQueue = new ArrayDeque<TreeNode>();
        nodeQueue.offer(root);
        boolean isOrderLeft = true;

        while (!nodeQueue.isEmpty()) {
            Deque<Integer> levelList = new LinkedList<Integer>();
            int size = nodeQueue.size();
            for (int i = 0; i < size; ++i) {
                TreeNode curNode = nodeQueue.poll();
                if (isOrderLeft) {
                    levelList.offerLast(curNode.val);
                } else {
                    levelList.offerFirst(curNode.val);
                }
                if (curNode.left != null) {
                    nodeQueue.offer(curNode.left);
                }
                if (curNode.right != null) {
                    nodeQueue.offer(curNode.right);
                }
            }
            ans.add(new LinkedList<Integer>(levelList));
            isOrderLeft = !isOrderLeft;
        }

        return ans;
    }
}

5. 岛屿数量

class Solution {

    public int[][] posztion = {
        {0,-1},{0,1},{1,0},{-1,0}
    };

    public boolean[][] isUsed = new boolean[301][301];

    public int numIslands(char[][] grid) {
        int ans = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (isUsed[i][j] == false && grid[i][j] == '1') {
                    bfs(grid,i,j,ans);
                    ans++;
                }
            }
        }
        return ans;
    }

    public void bfs(char[][] grid,int x,int y, int ans) {
        for (int i = 0; i < 4; i++) {
            int a = x + posztion[i][0];
            int b = y + posztion[i][1];
            if (a >= 0 && a < grid.length && b >= 0 && b < grid[0].length
                && isUsed[a][b] == false && grid[a][b] == '1') {
                    isUsed[a][b] = true;
                    bfs(grid,a,b,ans);
            }
        }
    }

}

6. 字典序排数（有难度）

原题链接

循环n次，一次只入一个数

class Solution {
    public List<Integer> lexicalOrder(int n) {
        List<Integer> ret = new ArrayList<Integer>();
        int number = 1;
        for (int i = 0; i < n; i++) {
            ret.add(number);
            if (number * 10 <= n) {
                number *= 10;
            } else {
                while (number % 10 == 9 || number + 1 > n) {
                    number /= 10;
                }
                number++;
            }
        }
        return ret;
    }
}

7. 克隆图

如果该节点已经被访问过了，则直接从哈希表中取出对应的克隆节点返回

原题链接

class Solution {
    private HashMap <Node, Node> visited = new HashMap <> ();
    public Node cloneGraph(Node node) {
        if (node == null) {
            return node;
        }

        // 如果该节点已经被访问过了，则直接从哈希表中取出对应的克隆节点返回
        if (visited.containsKey(node)) {
            return visited.get(node);
        }

        // 克隆节点，注意到为了深拷贝我们不会克隆它的邻居的列表
        Node cloneNode = new Node(node.val, new ArrayList());
        // 哈希表存储
        visited.put(node, cloneNode);

        // 遍历该节点的邻居并更新克隆节点的邻居列表
        for (Node neighbor: node.neighbors) {
            cloneNode.neighbors.add(cloneGraph(neighbor));
        }
        return cloneNode;
    }
}

8. 省份数量

原题链接

并查集

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int cities = isConnected.length;
        int[] parent = new int[cities];
        for (int i = 0; i < cities; i++) {
            parent[i] = i;
        }
        for (int i = 0; i < cities; i++) {
            for (int j = i + 1; j < cities; j++) {
                if (isConnected[i][j] == 1) {
                    union(parent, i, j);
                }
            }
        }
        int provinces = 0;
        for (int i = 0; i < cities; i++) {
            if (parent[i] == i) {
                provinces++;
            }
        }
        return provinces;
    }

    public void union(int[] parent, int index1, int index2) {
        parent[find(parent, index1)] = find(parent, index2);
    }

    public int find(int[] parent, int index) {
        if (parent[index] != index) {
            parent[index] = find(parent, parent[index]);
        }
        return parent[index];
    }
}

9. 课程表

原题链接

class Solution {

    private static final int N = 5001;
    private static int[] e;
    private static int[] ne;
    private static int[] h;
    private static int[] cnt;
    private static int idx;

    private static void add(int a,int b) {
        e[idx] = b;ne[idx] = h[a];cnt[b]++;h[a] = idx++;
    }

    public boolean canFinish(int numCourses, int[][] prerequisites) {
        e = new int[N];
        ne = new int[N];
        h = new int[N];
        cnt = new int[N];
        idx = 0;
        Arrays.fill(h,-1);
        for (int[] temp : prerequisites) {
            add(temp[1],temp[0]);
        }
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (cnt[i] == 0) {
                queue.add(i);
            }
        }
        int sum = 0;
        while (!queue.isEmpty()) {
            int r = queue.remove();
            sum++;
            for (int i = h[r]; i != -1; i = ne[i]) {
                int j = e[i];
                cnt[j]--;
                if (cnt[j] == 0) {
                    queue.add(j);
                }
            }
        }
        if (sum == numCourses) {
            return true;
        } else {
            return false;
        }
    }
}

10. 所有可能的路径

原题链接

class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    Deque<Integer> stack = new ArrayDeque<Integer>();

    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        stack.offerLast(0);
        dfs(graph, 0, graph.length - 1);
        return ans;
    }

    public void dfs(int[][] graph, int x, int n) {
        if (x == n) {
            ans.add(new ArrayList<Integer>(stack));
            return;
        }
        for (int y : graph[x]) {
            stack.offerLast(y);
            dfs(graph, y, n);
            stack.pollLast();
        }
    }
}

11. 判断二分图

原题链接

class Solution {
    private static final int UNCOLORED = 0;
    private static final int RED = 1;
    private static final int GREEN = 2;
    private int[] color;
    private boolean valid;

    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        valid = true;
        color = new int[n];
        Arrays.fill(color, UNCOLORED);
        for (int i = 0; i < n && valid; ++i) {
            if (color[i] == UNCOLORED) {
                dfs(i, RED, graph);
            }
        }
        return valid;
    }

    public void dfs(int node, int c, int[][] graph) {
        color[node] = c;
        int cNei = c == RED ? GREEN : RED;
        for (int neighbor : graph[node]) {
            if (color[neighbor] == UNCOLORED) {
                dfs(neighbor, cNei, graph);
                if (!valid) {
                    return;
                }
            } else if (color[neighbor] != cNei) {
                valid = false;
                return;
            }
        }
    }
}

12. 字符串解码

两个栈，一个存数字，一个存字母

res一直更新

class Solution {
    public String decodeString(String s) {
        StringBuilder res = new StringBuilder();
        int multi = 0;
        LinkedList<Integer> stack_multi = new LinkedList<>();
        LinkedList<String> stack_res = new LinkedList<>();
        for(Character c : s.toCharArray()) {
            if(c == '[') {
                stack_multi.addLast(multi);
                stack_res.addLast(res.toString());
                multi = 0;
                res = new StringBuilder();
            }
            else if(c == ']') {
                StringBuilder tmp = new StringBuilder();
                int cur_multi = stack_multi.removeLast();
                for(int i = 0; i < cur_multi; i++) tmp.append(res);
                res = new StringBuilder(stack_res.removeLast() + tmp);
            }
            else if(c >= '0' && c <= '9') multi = multi * 10 + Integer.parseInt(c + "");
            else res.append(c);
        }
        return res.toString();
    }
}

13. 快速排序

原题链接

class Solution {

    public int[] sortArray(int[] nums) {
        int n = nums.length;
        quickSort(nums,0,n-1);
        return nums;
    }

    public void quickSort(int[] nums, int l, int r) {
        
        if (l >= r) {
            return;
        }

        int i = l-1, j = r+1;
        int x = nums[i+j>>1];
        while (i < j) {
            do i++; while(nums[i] < x);
            do j--; while(nums[j] > x);
            if (i < j) {
                int temp = nums[i];
                nums[i] = nums[j];
                nums[j] = temp;
            }
        }
        quickSort(nums,l,j);
        quickSort(nums,j+1,r);
    }

}

14. 利用快速排序找到第k个数

原题链接

class Solution {
    public int findKthLargest(int[] nums, int k) {
        return quickSortFindKthLargest(nums,k,0,nums.length-1);
    }

    public int quickSortFindKthLargest(int[] nums, int k, int l,int r) {
        if (l >= r) {
            return nums[l];
        }

        int i = l-1, j = r+1;
        int x = nums[i+j>>1];
        while (i < j) {
            do i++; while(nums[i] < x);
            do j--; while(nums[j] > x);
            if (i < j) {
                int temp = nums[i];
                nums[i] = nums[j];
                nums[j] = temp;
            }
        }
        if (j+1<=nums.length-k) {
            return quickSortFindKthLargest(nums,k,j+1,r);
        } else {
            return quickSortFindKthLargest(nums,k,l,j);
        }

    }

}

15. 归并排序

原题链接

class Solution {
    int[] tempNums;
    public int[] sortArray(int[] nums) {
        tempNums = new int[nums.length];
        int n = nums.length;
        mergeSort(nums,0,n-1);
        return nums;
    }

    public void mergeSort(int[] nums, int l, int r) {
        if (l >= r) {
            return;
        }
        int mid = l+r>>1;
        mergeSort(nums,l,mid);
        mergeSort(nums,mid+1,r);
        int i = l,j = mid+1;
        int idx = 0;
        while (i <= mid && j <= r) {
            if (nums[i] <= nums[j]) {
                tempNums[idx++] = nums[i++];
            } else {
                tempNums[idx++] = nums[j++];
            }
        }
        while (i <= mid) {
            tempNums[idx++] = nums[i++];
        }
        while (j <= r) {
            tempNums[idx++] = nums[j++];
        }
        for (int q = l,k = 0; q <= r; q++) {
            nums[q] = tempNums[k++];
        }
    }

}

16. 逆序对的数量

原题链接

class Solution {

    public static int ans;
    
    public int reversePairs(int[] record) {
        ans = 0;
        int n = record.length;
        numberOfReverseOrderPairs(0,n-1,record);
        return ans;
    }

    public static void numberOfReverseOrderPairs(int l,int r,int[] nums){
        if (l >= r) {
            return;
        }
        int mid = (l+r)>>1;
        int i = l,j = mid+1;
        numberOfReverseOrderPairs(l,mid,nums);
        numberOfReverseOrderPairs(mid+1,r,nums);
        int[] temp = new int[r-l+1];
        int k = 0;
        while (i <= mid && j <= r) {
            if (nums[i] <= nums[j]) {
                temp[k++] = nums[i++];
            } else {
                temp[k++] = nums[j++];
                ans += mid - i + 1;
            }
        }
        while (i <= mid) {
            temp[k++] = nums[i++];
        }
        while (j <= r) {
            temp[k++] = nums[j++];
        }
        for (int q = l,p = 0; q <= r; q++) {
            nums[q] = temp[p++];
        }
    }

}

三、二分

17. 在排序数组中查找元素的第一个和最后一个位置

力扣链接
acwing链接

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return new int[]{-1,-1};
        }
        int l = 0, r = nums.length-1;
        while (l < r) {
            int mid = l+r>>1;
            if (nums[mid] < target) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        int ans1 = r;
        l = 0; r = nums.length-1;
        while (l < r) {
            int mid = l+r+1>>1;
            if (nums[mid] <= target) {
                l = mid;
            } else {
                r = mid-1;
            }
        }
        int ans2 = r;
        if (nums[ans1] == target && nums[ans2] == target) {
            return new int[]{ans1,ans2};
        } else {
            return new int[]{-1,-1};
        }
    }
}

18. 数的三次方根

力扣连接
acwing链接

class Solution {
    public int mySqrt(int x) {
        int l = 0, r = x, ans = -1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if ((long) mid * mid <= x) {
                ans = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return ans;
    }
}

四、高精度

19. 字符串相加

力扣链接
acwing链接

class Solution {
    public String addStrings(String num1, String num2) {
        int i = num1.length() - 1, j = num2.length() - 1, add = 0;
        StringBuffer ans = new StringBuffer();
        while (i >= 0 || j >= 0 || add != 0) {
            int x = i >= 0 ? num1.charAt(i) - '0' : 0;
            int y = j >= 0 ? num2.charAt(j) - '0' : 0;
            int result = x + y + add;
            ans.append(result % 10);
            add = result / 10;
            i--;
            j--;
        }
        // 计算完以后的答案需要翻转过来
        ans.reverse();
        return ans.toString();
    }
}

20. 高精度减法

原题链接

import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;

public class Main{
    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        String s1 = scanner.next();
        String s2 = scanner.next();
        List<Integer> A = new ArrayList<>();
        List<Integer> B = new ArrayList<>();
        for(int i = s1.length() - 1;i >= 0;i --) A.add(s1.charAt(i) - '0');
        for(int i = s2.length() - 1;i  >= 0; i --) B.add(s2.charAt(i) - '0');
        if(!cmp(A,B)){
            System.out.print("-");
        }
        List<Integer> C = sub(A,B);
        for(int i = C.size() - 1;i >= 0; i --){
            System.out.print(C.get(i));
        }


    }
    public static List<Integer> sub(List<Integer> A,List<Integer> B){
        if(!cmp(A,B)){
            return sub(B,A);
        }

        List<Integer> C = new ArrayList<>();
        int t = 0;
        for(int i = 0;i < A.size();i ++){
            //这里应该是A.get(i) - B.get(i) - t ，因为可能B为零，所以需要判断一下是不是存在
            t = A.get(i) - t;
            if(i < B.size()) t -= B.get(i);
            C.add((t + 10) % 10);

            if(t < 0) t = 1;
            else t = 0;
        }
        //删除指定下标下面的值
        while(C.size() > 1 && C.get(C.size() - 1) == 0)  C.remove(C.size() - 1);

        return C;
    }
    public static boolean cmp(List<Integer> A,List<Integer> B){
        if(A.size() != B.size()) return A.size() > B.size();
       /* if(A.size() >= B.size()){
            return true;
        }else{
            return false;
        }
        */
        for(int i = A.size() - 1;i >= 0;i --){
            if(A.get(i) != B.get(i)) {
                return A.get(i) > B.get(i);
            }
        }
        return true;
    }
}

21. 高精度乘法

二刷：

1. 在草稿纸上演算一遍 运算过程，便知道 代码逻辑

力扣链接
acwing链接

class Solution {
    public String multiply(String num1, String num2) {
        if (num1.equals("0") || num2.equals("0")) {
            return "0";
        }
        String ans = "0";
        int m = num1.length(), n = num2.length();
        for (int i = n - 1; i >= 0; i--) {
            StringBuffer curr = new StringBuffer();
            int add = 0;
            for (int j = n - 1; j > i; j--) {
                curr.append(0);
            }
            int y = num2.charAt(i) - '0';
            for (int j = m - 1; j >= 0; j--) {
                int x = num1.charAt(j) - '0';
                int product = x * y + add;
                curr.append(product % 10);
                add = product / 10;
            }
            if (add != 0) {
                curr.append(add % 10);
            }
            ans = addStrings(ans, curr.reverse().toString());
        }
        return ans;
    }

    public String addStrings(String num1, String num2) {
        int i = num1.length() - 1, j = num2.length() - 1, add = 0;
        StringBuffer ans = new StringBuffer();
        while (i >= 0 || j >= 0 || add != 0) {
            int x = i >= 0 ? num1.charAt(i) - '0' : 0;
            int y = j >= 0 ? num2.charAt(j) - '0' : 0;
            int result = x + y + add;
            ans.append(result % 10);
            add = result / 10;
            i--;
            j--;
        }
        ans.reverse();
        return ans.toString();
    }
}

22. 高精度除法

原题链接

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> div(vector<int> &A, int b, int &r)
{
    vector<int> C;
    r = 0;
    for (int i = A.size() - 1; i >= 0; i -- )
    {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    reverse(C.begin(), C.end());
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

int main()
{
    string a;
    vector<int> A;

    int B;
    cin >> a >> B;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');

    int r;
    auto C = div(A, B, r);

    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];

    cout << endl << r << endl;

    return 0;
}

五、前缀和S与差分a

23. 区域和检索 - 数组不可变

力扣链接
acwing链接

class NumArray {
    int[] sums;

    public NumArray(int[] nums) {
        int n = nums.length;
        sums = new int[n + 1];
        for (int i = 0; i < n; i++) {
            sums[i + 1] = sums[i] + nums[i];
        }
    }
    
    public int sumRange(int i, int j) {
        return sums[j + 1] - sums[i];
    }
}

24. 子矩阵的和

acwing链接
力扣链接

class NumMatrix {
    int[][] sums;

    public NumMatrix(int[][] matrix) {
        int m = matrix.length;
        if (m > 0) {
            int n = matrix[0].length;
            sums = new int[m + 1][n + 1];
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    sums[i + 1][j + 1] = sums[i][j + 1] + sums[i + 1][j] - sums[i][j] + matrix[i][j];
                }
            }
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        return sums[row2 + 1][col2 + 1] - sums[row1][col2 + 1] - sums[row2 + 1][col1] + sums[row1][col1];
    }
}

25. 差分

力扣链接
acwing链接

class Solution {
    public int[] corpFlightBookings(int[][] bookings, int n) {
        int[] nums = new int[n];
        for (int[] booking : bookings) {
            nums[booking[0] - 1] += booking[2];
            if (booking[1] < n) {
                nums[booking[1]] -= booking[2];
            }
        }
        for (int i = 1; i < n; i++) {
            nums[i] += nums[i - 1];
        }
        return nums;
    }
}

26. 差分矩阵

原题链接

import java.util.*;

public class Main {
    public static void main (String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int q = scanner.nextInt();
        int[][] splits = new int[n+2][m+2];
        int[][] sum = new int[n+2][m+2];
        for (int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                sum[i][j] = scanner.nextInt();
                splits[i][j] = sum[i][j] - sum[i-1][j] - sum[i][j-1] + sum[i-1][j-1];
            }
        }
        for (int i = 0; i < q; i++) {
            int x1 = scanner.nextInt();
            int y1 = scanner.nextInt();
            int x2 = scanner.nextInt();
            int y2 = scanner.nextInt();
            int c = scanner.nextInt();
            splits[x1][y1] += c;
            splits[x1][y2+1] -= c;
            splits[x2+1][y1] -= c;
            splits[x2+1][y2+1] += c;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                sum[i][j] = splits[i][j] + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];
                System.out.print(sum[i][j] + " ");
            }
            System.out.println();
        }
    }
}