力扣爆刷第87天之hot100五连刷21-25

一、240. 搜索二维矩阵 II

题目链接：https://leetcode.cn/problems/search-a-2d-matrix-ii/description/?envType=study-plan-v2&envId=top-100-liked
思路：从横向递增纵向递增的矩阵中搜索target，利用顺序的特性，从矩阵的右上角进行搜索，如果相等就返回，如果大于向下进行搜索，如果小于向左进行搜索，存在的话自然可以找到，不存在的话会一直搜索到边界才会停止，也就是当前算法还有改进的空间，可以采取早停措施。

class Solution {public boolean searchMatrix(int[][] matrix, int target) {return dfs(matrix, 0, matrix[0].length-1, target);}boolean dfs(int[][] matrix, int i, int j, int target) {while(i < matrix.length && j >= 0) {if(matrix[i][j] == target) {return true;}if(matrix[i][j] > target) {j--;}else {i++;}}return false;}
}

二、160. 相交链表

题目链接：https://leetcode.cn/problems/intersection-of-two-linked-lists/description/?envType=study-plan-v2&envId=top-100-liked
思路：经典题目，直接求长度，然后对齐，然后同步比较，相同即返回。

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode pa = headA, pb = headB;int lenA = 0, lenB = 0;while(pa != null) {lenA++;pa = pa.next;}while(pb != null) {lenB++;pb = pb.next;}pa = headA;pb = headB;while(lenA > lenB) {pa = pa.next;lenA--;}while(lenB > lenA) {pb = pb.next;lenB--;}while(pa != null) {if(pa == pb) {return pa;}pa = pa.next;pb = pb.next;}return null;}
}

三、206. 反转链表

题目链接：https://leetcode.cn/problems/reverse-linked-list/description/?envType=study-plan-v2&envId=top-100-liked
思路：两个指针，一个指向头结点，一个用于向后走，遍历过程中用临时变量记录临时节点，然后采用头插法即可。

class Solution {public ListNode reverseList(ListNode head) {ListNode root = new ListNode();ListNode p1 = root, p2 = head;while(p2 != null) {ListNode t = p2;p2 = p2.next;t.next = p1.next;p1.next = t;}return root.next;}
}

四、234. 回文链表

题目链接：https://leetcode.cn/problems/palindrome-linked-list/description/?envType=study-plan-v2&envId=top-100-liked
思路：直接快慢指针定位中间节点，然后翻转其中一般的链表，然后遍历比较即可。

/*** 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 boolean isPalindrome(ListNode head) {ListNode p1 = head, p2 = head;while(p2 != null && p2.next != null) {p1 = p1.next;p2 = p2.next.next;}if(p2 != null) {p1 = p1.next;}p2 = reverse(p1);p1 = head;while(p2 != null) {if(p1.val != p2.val) {return false;}p1 = p1.next;p2 = p2.next;}return true;}ListNode reverse(ListNode head) {ListNode root = new ListNode();ListNode p1 = root, p2 = head;while(p2 != null) {ListNode t = p2;p2 = p2.next;t.next = p1.next;p1.next = t;}return root.next;}
}

五、141. 环形链表

题目链接：https://leetcode.cn/problems/linked-list-cycle/description/?envType=study-plan-v2&envId=top-100-liked
思路：判断是否有环，直接快慢指针。

public class Solution {public boolean hasCycle(ListNode head) {ListNode slow = head, fast = head;while(fast != null && fast.next != null) {slow = slow.next;fast = fast.next.next;if(slow == fast) return true;}return false;}
}