[LeetCode]76.Minimum Window Substring，leetcode76.minimum

[LeetCode]76.Minimum Window Substring，leetcode76.minimum

題目

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.

分析

詳細些參考：[演算法系列之二十二]包含T全部元素的最小子視窗

代碼

/*--------------------------------------------*   日期：2015-02-24*   作者：SJF0115*   題目: 76.Minimum Window Substring*   網址：https://oj.leetcode.com/problems/minimum-window-substring/*   結果：AC*   來源：LeetCode*   總結：------------------------------------------------*/#include <iostream>#include <algorithm>#include <climits>using namespace std;class Solution {public:    string minWindow(string S, string T) {        int slen = S.size();        int tlen = T.size();        if(slen <= 0 || tlen <= 0){            return "";        }//if        int minWinStart = 0,minWinEnd = 0;        int minWinLen = INT_MAX;        // 儲存到目前為止遇到過的T中字元總數        int count = 0;        // 儲存T中不同字元的總數        int needFind[256] = {0};        for(int i = 0;i < tlen;++i){            ++needFind[T[i]];        }//for        // 儲存到目前為止遇到過的不同字元的總數        int hasFound[256] = {0};        int val;        for(int start = 0,end = 0;end < slen;++end){            val = S[end];            // 跳過不在T中的字元            if(needFind[val] == 0){                continue;            }//if            ++hasFound[val];            if(hasFound[val] <= needFind[val]){                ++count;            }//if            // 找到一個有效視窗            if(count == tlen){                int startVal = S[start];                while(needFind[startVal] == 0 ||                      hasFound[startVal] > needFind[startVal]){                    if(hasFound[startVal] > needFind[startVal]){                        --hasFound[startVal];                    }//if                    ++start;                    startVal = S[start];                }//while                // 更新最小視窗                int curWinLen = end - start + 1;                if(curWinLen < minWinLen){                    minWinLen = curWinLen;                    minWinStart = start;                    minWinEnd = end;                }//if            }//if        }//for        if(count != tlen){            return "";        }//if        return S.substr(minWinStart,minWinEnd - minWinStart + 1);    }};int main() {    Solution solution;    string S("acbbaca");    string T("aba");    cout<<solution.minWindow(S,T)<<endl;}

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