力扣爆刷第103天之CodeTop100五连刷1-5

一、3. 无重复字符的最长子串

题目链接：https://leetcode.cn/problems/longest-substring-without-repeating-characters/description/
思路：求最长无重复子串，典型滑动窗口，用set记录是否重复，无重扩大窗口，有重缩小窗口，每次添加元素后更新最大值。

class Solution {
  public int lengthOfLongestSubstring(String s) {
        int max = 0;
        Set<Character> set = new HashSet<>();
        int slow = 0;
        for(int i = 0; i < s.length(); i++) {
            while(set.contains(s.charAt(i))) {
                set.remove(s.charAt(slow));
                slow++;
            }
            set.add(s.charAt(i));
            max = Math.max(max, set.size());
        }
        return max;
    }
}

二、206. 反转链表

题目链接：https://leetcode.cn/problems/reverse-linked-list/description/
思路：翻转链表采用头插法，一个指针指向虚拟头结点保持不动，另一个指针每次向后走，进行头插。

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode root = new ListNode();
        ListNode p = head;
        while(p != null) {
            ListNode pre = p.next;
            p.next = root.next;
            root.next = p;
            p = pre;
        }
        return root.next;
    }
}

三、146. LRU 缓存

题目链接：https://leetcode.cn/problems/lru-cache/description/
思路：非常经典的题目，最近最少使用，我们需要构建一个双向链表和哈希表，其中双向链表应该有虚拟的头结点和尾结点，方便进行添加删除操作，其他的是常规操作。

class LRUCache {
    
    int capacity;
    Map<Integer, Node> map;
    DoubleList list;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        map = new HashMap<>();
        list = new DoubleList();
    }

    public int get(int key) {
        if(!map.containsKey(key)) return -1;
        Node temp = map.get(key);
        makeRecently(temp);
        return temp.value;
    }

    public void put(int key, int value) {
        if(map.containsKey(key)) {
            Node temp = map.get(key);
            temp.value = value;
            makeRecently(temp);
            return;
        }
        if(list.size == capacity) {
            Node temp = list.deleteFirst();
            map.remove(temp.key);
        }
        Node temp = new Node(key, value);
        list.add(temp);
        map.put(key, temp);
    }

    void makeRecently(Node temp) {
       list.delete(temp);
       list.add(temp);
    }
}

class DoubleList {

    int size = 0;
    Node start, end;

    DoubleList() {
        start = new Node();
        end = new Node();
        start.right = end;
        end.left = start;
    }

    void delete(Node temp) {
        temp.left.right = temp.right;
        temp.right.left = temp.left;
        temp.left = null;
        temp.right = null;
        size--;
    }

    Node deleteFirst() {
        if(start.right == end) return null;
        Node temp = start.right;
        delete(temp);
        return temp;
    }

    void add(Node temp) {
        end.left.right = temp;
        temp.left = end.left;
        temp.right = end;
        end.left = temp;
        size++;
    }
}

class Node{
    int key, value;
    Node left, right;
    Node(){}
    Node(int k, int v) {
        key = k;
        value = v;
    }
}

四、215. 数组中的第K个最大元素

题目链接：https://leetcode.cn/problems/kth-largest-element-in-an-array/description/
思路：使用快速排序超时，又要求时间复杂度为O(N)，使用桶排序最合数不过了。
桶排序：即把所有的数作为索引，放到一个新数组里去，即放到桶里，出现多次即累加，然后从后往前遍历减去次数，即可得到第K个最大的元素。

class Solution {
  public int findKthLargest(int[] nums, int k) {
        int[] count = new int[20001];
        for(int i : nums) {
            count[i + 10000]++;
        }
        for(int i = 20000; i > 0; i--) {
            k = k - count[i];
            if(k <= 0) return i - 10000; 
        }
        return 0;
    }
    
}

快排：

class Solution {
  public int findKthLargest(int[] nums, int k) {
        quickSort(nums, 0, nums.length-1);
        return nums[nums.length - k];
    }
    
    void quickSort(int[] nums, int left, int right) {
        if(left >= right) return;
        int base = nums[left];
        int i = left, j = right;
        while(i < j) {
            while(nums[j] >= base && i < j) j--;
            while(nums[i] <= base && i < j) i++;
            swap(nums, i, j);
        }
        swap(nums, left, j);
        quickSort(nums, left, j-1);
        quickSort(nums, j+1, right);
    }

    void swap(int[] nums, int x, int y) {
        int t = nums[x];
        nums[x] = nums[y];
        nums[y] = t;
    }
}

五、25. K 个一组翻转链表

题目链接：https://leetcode.cn/problems/reverse-nodes-in-k-group/description/
思路：

K个一组翻转，头插法、尾插法都可以，头插法需要虚拟头结点，尾插法不需要。
下面使用头插法。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode root = new ListNode(-1, head);
        ListNode slow = root, fast =head;
        int i = 0;
        while(fast != null) {
            i++;
            fast = fast.next;
            if(i % k == 0) {
                slow = reverse(slow, fast);
            }
        }
        return root.next;
    }

    ListNode reverse(ListNode a, ListNode b) {
        ListNode r = a, p1 = a.next, p2 = p1;
        r.next = null;
        while(p1 != b) {
            ListNode pre = p1.next;
            p1.next = r.next;
            r.next = p1;
            p1 = pre;
        }
        p2.next = b;
        return p2;
    }

   
}