[leetcode 76] Minimum Window Substring

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本文提供了一种复杂度为O(n)的算法，用于找到字符串S中包含所有目标字符串T字符的最短子串。通过双指针技巧实现，确保效率并解决相关问题。

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.

思路：最短摘要生成问题。双指针，尾指针不断往后走，直到该窗口包含了T中所有的字符，再讲头指针往后走，直到不能再走为止。

class Solution {
public:
    string minWindow(string S, string T) {
        const int ASCII_NUM = 256;
        const int n = S.size();
        const int m = T.size();
        if (n == 0 || n < m) {
            return "";
        }
        int expect_show[ASCII_NUM];
        int appear_show[ASCII_NUM];
        fill_n(&expect_show[0], ASCII_NUM, 0);
        fill_n(&appear_show[0], ASCII_NUM, 0);
        for (int i = 0; i < m; i++) {
            expect_show[T[i]]++;
        }
        int min_start = 0;
        int win_start = 0;
        int win_len = INT_MAX;
        int appear = 0;
        for (int win_end = 0; win_end < n; win_end++) {
            if (expect_show[S[win_end]] > 0) {
                appear_show[S[win_end]]++;
                if (appear_show[S[win_end]] <= expect_show[S[win_end]]) {
                    appear++;
                }
            }
            if (T.size() == appear) {
                while (appear_show[S[win_start]] > expect_show[S[win_start]] || expect_show[S[win_start]] == 0) {
                    appear_show[S[win_start]]--;
                    win_start++;
                }
                if (win_len > (win_end - win_start + 1)) {
                    win_len = win_end - win_start + 1;
                    min_start = win_start;
                }
            }
        }
        if (win_len == INT_MAX) {
            return "";
        }
        return S.substr(min_start, win_len);
        
    }
};