zoukankan      html  css  js  c++  java
  • Minimum Window Substring

    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 S that covers all characters in T, return the emtpy string "".

    If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

    解法:

    双指针,动态维护最小长度,记录最小长度的start 和 end.

    尾指针不断往后扫,当扫到有一个窗口包含了所有T的字符后,为指针继续向后扫,遇到在T中的字母时,check头指针是否能收缩,能收缩update最小长度,直到end到达S的end。

    public class Solution {
        public String minWindow(String S, String T) {
            
            if (S == null || T == null || S.length() == 0 || T.length() == 0) {
                return "";
            }
            int[] needToFind = new int[256];
            int[] hasFound = new int[256];
    
            for (int i = 0; i < T.length(); i++) {
                needToFind[T.charAt(i)]++;
            }
    
            int minWinLen = Integer.MAX_VALUE;
            int count = 0, tLen = T.length();
            int winBeg = 0, winEnd = 0;
            for (int begin = 0, end = 0; end < S.length(); end++) {
                if (needToFind[S.charAt(end)] == 0) {
                    continue;
                }
                hasFound[S.charAt(end)]++;
                if(hasFound[S.charAt(end)] <= needToFind[S.charAt(end)]){
                    count ++;
                }
                
                if(count == tLen){
                   //当begin所指的元素不在T中 或  之后又出现了beigin所指的元素 begin向后移动
                    while(needToFind[S.charAt(begin)] == 0 || hasFound[S.charAt(begin)] > needToFind[S.charAt(begin)]){
                        if(hasFound[S.charAt(begin)] > needToFind[S.charAt(begin)]){
                            //之后又出现了beigin所指的元素 begin向后移动 重复出现的元素个数减1
                            hasFound[S.charAt(begin)]--;
                        }
                        begin ++;
                    }
                    
                    int winLen = end - begin + 1;
                    if(winLen < minWinLen){
                        winBeg = begin;
                        winEnd = end;
                        minWinLen = winLen;
                    }
                }
            }
    
            if (count == T.length()) {
                return S.substring(winBeg, winEnd + 1);
            }
    
            return "";
        }
    }
  • 相关阅读:
    LOL 战斗力查询
    D3js-对柱状图的增,删,排序
    我的项目7 js 实现歌词同步(额,小小的效果)
    为什么电脑启动任务管理器会这样
    OpenCV求取轮廓线
    leetcode-Reverse Words in a String
    Linux lvs DR配置
    p2p网贷3种运营模式
    T4308 数据结构判断
    1080 线段树练习
  • 原文地址:https://www.cnblogs.com/RazerLu/p/3545270.html
Copyright © 2011-2022 走看看