Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example,S ="ADOBECODEBANC"T ="ABC" Minimum window is"BANC". Note: If there is no such window in…
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example,S = "ADOBECODEBANC"T = "ABC" Minimum window is "BANC". Note:If there is no such window i…
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BANC" Note: If there is no such window in S…
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example,S = "ADOBECODEBANC"T = "ABC" Minimum window is "BANC". Note:If there is no such window i…
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BANC" Note: If there is no such window in S…
Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W. If there is no such window in S that covers all characters in T, return the empty string "". If there are multiple such minimum-length window…
原题地址:https://oj.leetcode.com/problems/minimum-window-substring/ 题意: Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example,S = "ADOBECODEBANC"T = "ABC" M…
题目: Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example,S = "ADOBECODEBANC"T = "ABC" Minimum window is "BANC". Note:If there is no such wind…
题目 最小子串覆盖 给定一个字符串source和一个目标字符串target,在字符串source中找到包括所有目标字符串字母的子串. 样例 给出source = "ADOBECODEBANC",target = "ABC" 满足要求的解  "BANC" 注意 如果在source中没有这样的子串,返回"",如果有多个这样的子串,返回起始位置最小的子串. 挑战 要求时间复杂度为O(n) 说明 在答案的子串中的字母在目标字符串中是否…
我的超时思路,感觉自己上了一个新的台阶,虽然超时了,但起码是给出了一个方法. 遍历s 一遍即可,两个指针,当找到了一个合格的字串后,start 开始走,直到遇到s[start]在t中 如果不符合,end++,直到遇到s[end]在t中. class Solution {public: string minWindow(string s, string t) { int start=0,end=t.size()-1; string res; int len=INT_MAX; map<char,in…