算法刷题笔记 3.25-3.29
- 1. 相同的树
- 2. 二叉树的最近公共祖先
- 3. 二叉搜索树中第K小的元素
- 通过双端队列duque 中序遍历
- 4. 二叉树的锯齿形层序遍历
- new LinkedList<Integer>(levelList)双端队列复制 数组
- 需要左右顺序，考虑双端队列
- 5. 岛屿数量
- 6. 字典序排数（有难度）
- 循环n次，一次只入一个数
- 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
- 8. 省份数量
- 并查集
- 9. 课程表
- 10. 所有可能的路径
- 11. 判断二分图
- 12. 字符串解码
- 两个栈，一个存数字，一个存字母
- res一直更新
- 13. 快速排序
- 14. 利用快速排序找到第k个数
- 15. 归并排序
- 16. 逆序对的数量
- 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
- 18. 数的三次方根
- 四、高精度
- 19. 字符串相加
- 20. 高精度减法
- 21. 高精度乘法
- 22. 高精度除法
- 五、前缀和S与差分a
- 23. 区域和检索 - 数组不可变
- 24. 子矩阵的和
- 25. 差分
- 26. 差分矩阵
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：
- 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();
}
}
}
