Hust oj 1758 Minimum Window Substring(最小覆盖子串问题）

Minimum Window Substring

Time Limit: 1000 MS Memory Limit: 65535 K Total Submit: 42(15 users) Total Accepted: 27(13 users) Rating: Special Judge: No

Description

Given a string S and a string T, find the minimum window in S which will contain all the characters. For example, S = "ADOBECODEBANC" T = "ABC" Minimum window is "BANC". Note: If there are multiple such windows, you are guaranteed the window with the shortest length and first appeared in the string S.

Input

There are multiple test cases, processing to the end-of-file. For each case: Line 1: This line contains string S. Line 2: This line contains string T. The string contains printable characters except space (visible characters). The length of the string is not less than 1 and not greater than 100,000.

Output

For each case: Output one line, contains the minimum window substring. If there is no such window in S that covers all characters in T, output the empty string "".

Sample Input

ADOBECODEBANC ABC acbdbaab aabd adobecodebancbbcaa abc 203838534736761996179497200837 746 ZHEXOMNHPRWMAJFPSUCYFZXUNLNRRU YUX dptpeaevzsykxnfurfbqcxdqkujgtc urn uLY7JdYK1810x95bT3LyPx7V37iLjY Jx8 w_J)u^^@gsWoXnTB*@CWAh./k9d3|% ^3w

Sample Output

BANC dbaa bca 4736 YFZXU nfur JdYK1810x w_J)u^^@gsWoXnTB*@CWAh./k9d3

LeetCode上的一道题，从别人那里学到了O(n)复杂度的算法，感觉很巧妙，维护一个合法的窗口，不断缩小这个窗口来达到求最短字符串的目的，双指针加贪心，具体的可以看这篇博客：http://www.cnblogs.com/lichen782/p/leetcode_minimum_window_substring_3.html

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int Maxn = 100005;
char str1[Maxn];
char str2[Maxn];
int FirMap[260];
int SecMap[260];
int tempStart,tempEnd;
int Start,End;
int MinLength;

int main()
{
    while(~scanf("%s",&str1))
    {
        scanf("%s",&str2);
        memset(FirMap,0,sizeof(FirMap));
        memset(SecMap,0,sizeof(SecMap));
        int len_1 = strlen(str1);
        int len_2 = strlen(str2);
        for(int i=0;i<len_2;i++)
        {
            FirMap[str2[i]]++;
        }
        int ans = 0;
        MinLength = len_1;
        End = len_1;
        Start = -1;
        tempStart = 0;
        int i;
        for(int i=0;i<len_1;i++)
        {
            SecMap[str1[i]]++;
            if(SecMap[str1[i]] <= FirMap[str1[i]])
            {
                ans++;
            }
            if(ans == len_2)
            {
                while(tempStart < i && SecMap[str1[tempStart]] > FirMap[str1[tempStart]])
                {
                    SecMap[str1[tempStart]]--;
                    tempStart++;
                }

                if(i - tempStart < MinLength)
                {
                    MinLength = i - tempStart;
                    End = i;
                    Start = tempStart;
                }
                SecMap[str1[tempStart]]--;
                ans--;
                tempStart++;
            }
        }
           if(Start != -1)
                for(int i=Start;i<=End;i++)
                {
                    printf("%c",str1[i]);
                }
        printf("\n");
    }
}