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). 例子 Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BANC" 解决方案 ** Solution Java Method one ** ** 3ms, beats 92.14% ** ** 39.8MB...
速度还算可以。 publicclassSolution {publicbooleancover(int[] a,int[] b){for(inti = 0;i<256;i++)if( a[i] <b[i])returnfalse;returntrue; }publicString minWindow(String s, String t) {intlen1 =s.length();intlen2 =t.length();int[] pos =newint[256];for(inti = 0;i<len2;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()) { return ...
python java c++ 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...
LeetCode Top 100 Liked Questions 76. Minimum Window Substring (Java版; Hard) 题目描述 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" ...
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...
1publicclassSolution {2publicString minWindow(String S, String T) {3//Start typing your Java solution below4//DO NOT write main() function5if(S ==null|| S.length() == 0 || T ==null|| T.length() == 0)return"";6HashMap<Integer, Integer> map =newHashMap<Integer, Integer>()...
Minimum Window Substring 欢迎访问原文所在博客:https://52heartz.top/articles/leetcode-76-minimum-window-substring/ 解答1[Java]: 思路 注意 if (map[s.charAt(right++)]-- > 0) 这句,不论 right 指向位置的字符是哪个,在判断之后都会被减去 1。因为在前边的语句中,已经预先把 t 中的字符在 map 中...
publicclassMinimumWindowSubstring{ /** * ac */ publicStringminWindow2(Strings,Stringt) { if(s==null||s.length() ==0||t==null||t.length() ==0) { return""; } intcnt=0; int[]a=newint[256]; for(inti=0;i<t.length();i++) { ...
下一个子串了start++;}}// 如果begin没有修改过,返回空returnbegin==-1?"":S.substring(begin,end+1);}} 解题思路其实就是通过双指针维持一个Window,窗口右指针向右扩张用来找到包含子串为目的,窗口左指针向右收缩以使子串最小。 典型的滑动窗口方法的实现。