LeetCode 76. Minimum Window Substring (双指针)

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本文深入探讨了在复杂度为O(n)下寻找包含所有目标字符的最小子串问题，通过具体实例“ADOBECODEBANC”与“ABC”，详细解释了一种高效算法的实现过程。该算法首先构建目标字符串的字符频率映射，然后遍历源字符串，动态调整窗口大小，直至找到符合条件的最短子串。

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76. Minimum Window Substring

Hard

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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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

If there is no such window in S that covers all characters in T, return the empty string "".
If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

题意：

题解：

class Solution {
public:
    bool is_window_ok(int map_s[], int map_t[], vector<int>& vec_t){
        for (int i = 0; i < vec_t.size(); i++){
            if(map_s[vec_t[i]] < map_t[vec_t[i]]){
                return false;
            }
        }
        return true;
    }
    string minWindow(string s, string t) {
        const int MAX_ARRAY_LEN = 128;
        int map_t[MAX_ARRAY_LEN] = {0};
        int map_s[MAX_ARRAY_LEN] = {0};
        vector<int> vec_t;
        for (int i = 0; i < t.length(); i++){
            map_t[t[i]]++;
        }
        for (int i = 0; i < MAX_ARRAY_LEN; i++){
            if(map_t[i] > 0){
                vec_t.push_back(i);
            }
        }
        int window_begin = 0;
        string result = "";
        for (int i = 0; i < s.length(); i++){
            map_s[s[i]]++;
            while(window_begin < i){
                if(map_t[s[window_begin]] == 0){
                    window_begin++;
                } else if(map_s[s[window_begin]] > map_t[s[window_begin]]){
                    map_s[s[window_begin]]--;
                    window_begin++;
                } else {
                    break;
                }
            }
            if(is_window_ok(map_s, map_t, vec_t)){
                if(result == "" || result.size() > (i - window_begin + 1)){
                    result = s.substr(window_begin, i - window_begin + 1);
                }
            }
        }
        return result;
    }
};