class Solution: def minWindow(self, s: str, t: str) -> str: m: int = len(s) # count[ch] 表示滑动窗口 [l, r) 中字母 ch 还需出现的次数。 # 初始化为 t 中每个字母的出现次数 count: Dict[int, int] = Counter(t) # remain 表示滑动窗口 [l, r) 中还需出现的不同字母数 remain...
Lintcode32 Minimum Window Substring solution 题解 【题目描述】 Given a string source and a string target, find the minimum window in source which will contain all the characters in target. Notice:If there is no such window in source that covers all characters in target, return the emtpy stri...
C++代码实现: #include<iostream>#include<string>usingnamespacestd;classSolution {public:stringminWindow(stringS,stringT) {intlens=S.length();intlent=T.length();inttcnt[256]={0};intfcnt[256]={0};inti,start;intfound=0;intwinStart=-1,winEnd=lens;for(i=0;i<lent;i++) tcnt[T[i]]++...
minimum window in S. Solution: 1. Use two pointers: start and end. First, move 'end'. After finding all the needed characters, move 'start'. 2. Use array as hashtable. 1classSolution {2public:3stringminWindow(stringS,stringT) {4intN = S.size(), M =T.size();5if(N < M)retur...
from collections import defaultdict class Solution: """ @param source : A string @param target: A string @return: A string denote the minimum window, return "" if there is no such a string """ def minWindow(self, source , target): # 初始化counter_s和counter_t counter_s = defaultdict...
# 滑动窗口法 import collections class Solution: def minWindow(self, s, t): count = collections.Counter(t) # 统计字符串 t 中字符出现的次数 miss = len(t) # miss 负责记录 当前窗口是否满足条件 i = m = n = 0 for j, v in enumerate(s, 1): # 向右滑动或扩展 miss -= (count[v] ...
public class Solution { public String minWindow(String S, String T) { if (S == null || S.length() < 1 || T == null || T.length() < 1) { return ""; } int[] n = new int[128]; for (char c : T.toCharArray()) { n[c]++; } int end = 0, begin = 0, d = Inte...
class Solution { 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) { ...
public class Solution { public String minWindow(String s, String t) { // 建map, 记录被包含字符串中字符及其个数 Map<Character, Integer> map = new HashMap<>(); for (int i = 0; i < t.length(); i++) { char c = t.charAt(i); ...
https://leetcode.cn/problems/minimum-window-substring/ 一、个人版本 public class Solution1 { public String minWindow(String s, String t) { Balance balance = new Balance(s, t); Integer[] idxList = balance.idxList; if (idxList.length == 0 || idxList.length < t.length()) { ...