2019-12-13 20:19 − The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any ... 57one 0 251 [LeetCode]461. Hamming Distance 2019-12-23 16:39 − The Hamming distance between two integers ...
状态转移方程借助53. Edit Distance between strings上的这张图片来说明: LeetCode那道题的实现代码,时间复杂度 O(n*n),空间复杂度O(n) classSolution {public:intminDistance(stringword1,stringword2) {if(word1.empty() && word2.empty())return0;intlen1 = word1.length(), len2 =word2.length();...
LeetCode力扣 871. 最低加油次数 Minimum Number of Refueling Stops 18 -- 5:54 App LeetCode力扣 990. 等式方程的可满足性 Satisfiability of Equality Equations 18 -- 6:17 App LeetCode力扣 161. 相隔为 1 的编辑距离 One Edit Distance 168 -- 16:17 App LeetCode力扣 47. 全排列 II Permutation...
1. 一些定义 Minimum Edit Distance 编辑距离(Edit distance)帮助我们度量两个字符的相似程度。最小...Minimum Area Rectangle II Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes....
Input: s = "ADOBECODEBANC", t = "ABC" Output: "BANC" Example 2: Input: s = "a", t = "a" Output: "a" 通过滑动窗口法(sliding-window)求解。定义变量left和right,match,定义数组map。其中left和right为s的子串的左右边界,left,right初始化为0。start为满足条件的最短子串的起始位置。len为满足...
LeetCode 712. Minimum ASCII Delete Sum for Two Strings Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal. kind of like edit distance and the LC583. now, we are not asking to delete as less characters as po......
输入:s1 = "delete", s2 = "leet"输出:403解释:在 "delete" 中删除 "dee" 字符串变成 "let", 将 100[d]+101[e]+101[e] 加入总和。在 "leet" 中删除 "e" 将 101[e] 加入总和。 结束时,两个字符串都等于 "let",结果即为 100+101+101+101 = 403 。 如果改为将两个字符串转换为 "lee"...
Minimum Window Substring@LeetCode Minimum Window Substring 典型的窗口操作题,维护两个哈希表,stdMap标准表,map当前表,标准表用来保存T中的字符信息,当前表用来保存当前窗口的字符信息。 对窗口的操作包括以下两个: 扩充窗口:将窗口的右端点尽力向右扩展,直至到包含所有标准表中的字符(窗口中的每个有效字符的数量...
public: string minWindow(string s, string t) { string res = ""; int Cnt = 0; int left = 0; int minLen = INT_MAX; unordered_map<char, int>letterCnt; for (char c : t)++letterCnt[c]; for (int i = 0; i < s.size(); ++i) { ...
while(right < nums.size()){ while(right < nums.size() && sum < s){//扩大窗口 sum += nums[right++]; } while(sum >= s){//缩小窗口 res = min(res, right-left); sum -= nums[left++]; } } returnres == nums.size() +1?0: res; } };...