力扣爆刷第118天之CodeTop100五连刷76-80

一、221. 最大正方形

题目链接：https://leetcode.cn/problems/maximal-square/description/
思路：求最大的正方形，可以定义dp[i][j]表示nums[i][j]=1时，且以它为右下角的正方形的最长边长，有了定义以后就要想办法找出状态之间的转移，右下角是依赖于左边，上边，左上角的，关系如何传递？右下角的边长应该是这三个边长中最短的加1。由此便可以达到递推公式：dp[i+1][j+1] = Math.min(Math.min(dp[i+1][j], dp[i][j+1]), dp[i][j]) + 1;

class Solution {public int maximalSquare(char[][] matrix) {int m = matrix.length, n = matrix[0].length, max = matrix[0][0] - '0';int[][] dp = new int[m+1][n+1];for(int i = 0; i < m; i++) {for(int j = 0; j < n; j++) {if(matrix[i][j] == '0') continue;if(i == 0 || j == 0) {dp[i+1][j+1] = 1;}else{dp[i+1][j+1] = Math.min(Math.min(dp[i+1][j], dp[i][j+1]), dp[i][j]) + 1;}max = Math.max(max, dp[i+1][j+1]);}}return max * max;}
}

二、240. 搜索二维矩阵 II

题目链接：https://leetcode.cn/problems/search-a-2d-matrix-ii/description/
思路：矩阵本身是从左往右递增，从上往下递增，所以要利用这个特性，当前位置的左边都是比它小的，当前位置的下边都是比它大的，从右上角开始向左或者向下寻找，深度优先递归即可。

class Solution {boolean flag = false;public boolean searchMatrix(int[][] matrix, int target) {dfs(matrix, target, 0, matrix[0].length-1);return flag;}void dfs(int[][] matrix, int target, int x, int y) {if(x < 0 || x >= matrix.length || y < 0 || y >= matrix[0].length || flag) return;if(matrix[x][y] == target) {flag = true;return;}else if(target < matrix[x][y]) {dfs(matrix, target, x, y-1);}else{dfs(matrix, target, x+1, y);}}}

三、162. 寻找峰值

题目链接：https://leetcode.cn/problems/find-peak-element/description/
思路：题目要求时间复杂度log n，显然要使用二分查找，不过本题的二分查找因为没有一个明确的目的，是要去左区间还是右区间，那就是全部都要去，都要遍历，所以就是一个二叉树的遍历，相当于要前序遍历二叉树，剩下的就是判断和边界条件了，以及早停。

class Solution {int flag = -1;public int findPeakElement(int[] nums) {if(nums.length == 1) return 0;bs(nums, 0, nums.length-1);return flag;}void bs(int[] nums, int left, int right) {if(left > right || flag != -1) return ;int mid = left + (right - left) / 2;if(mid-1 == -1) {if(nums[mid] > nums[mid+1]) {flag = mid;return ;}}else if(mid+1 == nums.length) {if(nums[mid] > nums[mid-1]) {flag = mid;return ;}}else if(nums[mid] > nums[mid-1] && nums[mid] > nums[mid+1]) {flag = mid;return ;}bs(nums, left, mid-1);bs(nums, mid+1, right);}
}

四、234. 回文链表

题目链接：https://leetcode.cn/problems/palindrome-linked-list/description/
思路：链表求回文，可以快慢指针找到链表中间节点，然后翻转后半部分，然后比较，另外就是要注意链表的奇偶个数，要做一下处理。

class Solution {public boolean isPalindrome(ListNode head) {ListNode slow = head, fast = head;while(fast != null && fast.next != null) {slow = slow.next;fast = fast.next.next;}if(fast != null) slow = slow.next;fast = reverse(slow);slow = head;while(fast != null) {if(slow.val != fast.val) return false;slow = slow.next;fast = fast.next;}return true;}ListNode reverse(ListNode node) {ListNode root = new ListNode();ListNode p = node, pre = null;while(p != null) {pre = p.next;p.next = root.next;root.next = p;p = pre;}return root.next;}
}

五、112. 路径总和

题目链接：https://leetcode.cn/problems/path-sum/description/
思路：路径总和很简单，就是在二叉树中使用回溯算法，思路和典型的回溯类似，确定好收集结果的点，前序遍历，当前节点左右子树遍历完，向上返回时要去掉前面添加的状态。

class Solution {boolean flag = false;int sum = 0;public boolean hasPathSum(TreeNode root, int targetSum) {order(root, targetSum);return flag;}void order(TreeNode root, int targetSum) {if(root == null || flag) return;sum += root.val;if(root.left == null && root.right == null) {if(sum == targetSum){flag = true;return;}}order(root.left, targetSum);order(root.right, targetSum);sum -= root.val;}
}