随笔分类 -  LeetCode题解

摘要:1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), 阅读全文
posted @ 2020-03-24 12:14 Jinxiaobo0509 阅读(137) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), 阅读全文
posted @ 2020-03-24 12:04 Jinxiaobo0509 阅读(145) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), 阅读全文
posted @ 2020-03-24 11:24 Jinxiaobo0509 阅读(153) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), 阅读全文
posted @ 2020-03-23 22:25 Jinxiaobo0509 阅读(123) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), 阅读全文
posted @ 2020-03-23 21:04 Jinxiaobo0509 阅读(150) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), 阅读全文
posted @ 2020-03-23 17:36 Jinxiaobo0509 阅读(132) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), 阅读全文
posted @ 2020-03-23 15:52 Jinxiaobo0509 阅读(123) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 阅读全文
posted @ 2020-03-23 15:09 Jinxiaobo0509 阅读(138) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 阅读全文
posted @ 2020-03-22 22:10 Jinxiaobo0509 阅读(147) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 阅读全文
posted @ 2020-03-22 21:47 Jinxiaobo0509 阅读(137) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 阅读全文
posted @ 2020-03-22 21:03 Jinxiaobo0509 阅读(107) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 阅读全文
posted @ 2020-03-22 17:54 Jinxiaobo0509 阅读(86) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 阅读全文
posted @ 2020-03-21 23:44 Jinxiaobo0509 阅读(109) 评论(0) 推荐(0)
摘要:1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 阅读全文
posted @ 2020-03-21 22:45 Jinxiaobo0509 阅读(98) 评论(0) 推荐(0)
摘要:1 class Solution 2 { 3 public: 4 int hIndex(vector<int>& nums) 5 { 6 int l = 0,r = nums.size(); 7 while(l < r) 8 { 9 int mid = l + r + 1 >> 1; 10 11 / 阅读全文
posted @ 2020-03-21 21:42 Jinxiaobo0509 阅读(97) 评论(0) 推荐(0)
摘要:1 class Solution 2 { 3 public: 4 int findPeakElement(vector<int>& nums) 5 { 6 int l = 0,r = nums.size() - 1; 7 while(l < r) 8 { 9 int mid = l + r >> 1 阅读全文
posted @ 2020-03-21 20:30 Jinxiaobo0509 阅读(121) 评论(0) 推荐(0)
摘要:1 // Forward declaration of isBadVersion API. 2 bool isBadVersion(int version); 3 4 class Solution 5 { 6 public: 7 int firstBadVersion(int n) 8 { 9 in 阅读全文
posted @ 2020-03-21 20:02 Jinxiaobo0509 阅读(112) 评论(0) 推荐(0)
摘要:1 //利用模板 2 3 //以最后一个值来划分 4 //如果比它小,在右边,更新r 5 //反之,更新l 6 class Solution 7 { 8 public: 9 int findMin(vector<int>& nums) 10 { 11 int l = 0,r = nums.size( 阅读全文
posted @ 2020-03-21 19:25 Jinxiaobo0509 阅读(127) 评论(0) 推荐(0)
摘要:1 class Solution 2 { 3 public: 4 bool searchMatrix(vector<vector<int>>& matrix, int target) 5 { 6 if(matrix.empty() || matrix[0].empty()) return false 阅读全文
posted @ 2020-03-21 18:24 Jinxiaobo0509 阅读(149) 评论(0) 推荐(0)
摘要:1 使用数组中的值作为索引下标进行遍历,遍历的结果肯定是一个环(有一个重复元素) 2 检测重复元素问题转换成检测环的入口 3 为了找到环的入口,可以进行如下步骤: 4 5 设置两个快慢指针, fast每次走两步,slow每次走一步,最终走了slow走了n步与fast相遇,fast走了2*n,fast 阅读全文
posted @ 2020-03-21 17:05 Jinxiaobo0509 阅读(139) 评论(0) 推荐(0)