随笔分类 -  # 数据结构

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< 阅读全文
posted @ 2019-12-25 13:24 repinkply 阅读(7) 评论(0) 推荐(0)
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 阅读全文
posted @ 2019-12-23 12:52 repinkply 阅读(6) 评论(0) 推荐(0)
摘要:/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), r 阅读全文
posted @ 2019-12-23 11:38 repinkply 阅读(7) 评论(0) 推荐(0)
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; 阅读全文
posted @ 2019-12-22 22:00 repinkply 阅读(4) 评论(0) 推荐(0)
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 阅读全文
posted @ 2019-12-22 21:18 repinkply 阅读(8) 评论(0) 推荐(0)
摘要:第一:选择排序 #include <iostream> #include <string> using namespace std; template <typename T> void Swap(T& a,T& b) { T c=a; a=b; b=c; } template <typename 阅读全文
posted @ 2019-10-02 10:41 repinkply 阅读(10) 评论(0) 推荐(0)