LeetCode[Hash Table]: 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.

对于这个题目，我毫无头绪，当我在Discuss上看懂这个解法的时候：https://oj.leetcode.com/discuss/10337/accepted-o-n-solution我觉得这个解法实在是太妙了！这个链接有关于这个解法的idea：https://oj.leetcode.com/discuss/5469/is-the-length-of-t-considered-constant-or-m
下面先贴出这个解法的代码：

string minWindow(string S, string T) {
        int count = T.size();

        int  require[128] = { 0 };
        bool charSet[128] = { false };
        for (int n = 0; n < count; ++n) {
            require[T
]++;
            charSet[T
] = true;
        }

        int i = -1, j = 0;
        int minLen = INT_MAX, minIdx = 0;
        while (i < (int)S.size() && j < (int)S.size()) {
            if (count) {
                i++;
                require[S[i]]--;
                if (charSet[S[i]] && require[S[i]] >= 0) count--;
            }
            else {
                if (minLen > i - j + 1) {
                    minLen = i - j + 1;
                    minIdx = j;
                }
                require[S[j]]++;
                if (charSet[S[j]] && require[S[j]] > 0) count++;
                j++;
            }
        }

        return (minLen == INT_MAX ? "" : S.substr(minIdx, minLen));
    }

直观地理解这个算法是比较困难的，建议用一些例子跑起来，这样会比较容易理解这个算法为什么能到期望的目的。总之我很为这个解法而惊叹！
这个算法的时间性能表现如下图所示：