LeetCode 题解（101）: Minimum Window Substring

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 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.

题解：

双指针保持count == t.length()的移动。

c++版：

class Solution {public:string minWindow(string s, string t) {if (s.length() < t.length())return "";int count = 0;bool found = false;int start = 0, end = 0;int distance = INT_MAX;string result ="";int needToFind[256] = { 0 }, hasFound[256] = {0};for (int i = 0; i < t.length(); i++)needToFind[t[i]]++;while (end < s.length()) {if (needToFind[s[end]] == 0) {end++;continue;}hasFound[s[end]]++;if (hasFound[s[end]] <= needToFind[s[end]])count++;if (count == t.length()) {while (needToFind[s[start]] == 0 || hasFound[s[start]] > needToFind[s[start]]) {if (hasFound[s[start]] > needToFind[s[start]])hasFound[s[start]]--;start++;}if (end - start + 1 < distance) {distance = end - start + 1;result = s.substr(start, distance);}}end++;}return result;}};

Java版：

public class Solution {    public String minWindow(String s, String t) {        if(s.length() < t.length())            return "";                int[] hasFound = new int[256];        int[] needToFind = new int[256];        for(int i = 0; i < t.length(); i++) {            needToFind[t.charAt(i)]++;        }                int distance = Integer.MAX_VALUE;        int start = 0, end = 0;        int count = 0;        String result = "";        while(end < s.length()) {            if(needToFind[s.charAt(end)] == 0) {                end++;                continue;            }            hasFound[s.charAt(end)]++;            if(hasFound[s.charAt(end)] <= needToFind[s.charAt(end)])                count++;            if(count == t.length()) {                while(needToFind[s.charAt(start)] == 0 || hasFound[s.charAt(start)] > needToFind[s.charAt(start)]) {                    if(hasFound[s.charAt(start)] > needToFind[s.charAt(start)])                        hasFound[s.charAt(start)]--;                    start++;                }                if(end - start + 1 < distance) {                    distance = end - start + 1;                    result = s.substring(start, end + 1);                }            }            end++;        }        return result;    }}

Python版：

import sysclass Solution:    # @param {string} s    # @param {string} t    # @return {string}    def minWindow(self, s, t):        needToFind, hasFound, start, end, count, result, distance = [0] * 256, [0] * 256, 0, 0, 0, "", sys.maxint        for i in t:            needToFind[ord(i)] += 1        while end < len(s):            if needToFind[ord(s[end])] == 0:                end += 1                continue            hasFound[ord(s[end])] += 1            if hasFound[ord(s[end])] <= needToFind[ord(s[end])]:                count += 1            if count == len(t):                while needToFind[ord(s[start])] == 0 or hasFound[ord(s[start])] > needToFind[ord(s[start])]:                    if hasFound[ord(s[start])] > needToFind[ord(s[start])]:                        hasFound[ord(s[start])] -= 1                    start += 1                if end - start + 1 < distance:                    distance = end - start + 1                    result = s[start: end+1]            end += 1        return result