12 2019 档案
摘要://Time:(O(log(n))),Space:O(1) //此题为二分查找的变形题目,需要先学会二分查找法 class Solution { public: int binarySearchLast(vector<int>& nums,int target) { int begin=0; int
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摘要://解法一: class Solution { public: int firstUniqChar(string s) { int ret=-1; int a[26]={0}; for(int i=0;i<s.size();i++) { a[s[i]-'a']++; } for(int i=0;i<
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摘要://解法一:非递归版,最常见的一种二分查找算法 class Solution { public: int search(vector<int>& nums, int target) { int ret=-1; int begin=0; int end=nums.size()-1; while(beg
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摘要://解法一:递归版算法 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), l
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摘要:/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), r
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摘要://中序遍历二叉树,解法一:递归版 //Time:O(n) SPace:O(n) /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right;
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摘要://解法一:非递归版本 Time:O(n),Space:O(n) /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNo
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