字符串 --- 15.6 Minimum Window Substring --- 图解

本文介绍了一种在字符串S中找到包含字符串T所有字符的最小子串的方法，使用O(n)复杂度。通过实例演示了如何从ADOBECODEBANC中找出覆盖ABC的最小窗口BANC。文章详细解释了算法步骤，包括初始化计数数组、遍历字符串、调整窗口大小等，确保理解并实现高效解决方案。

/**********************************************************************************************************
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
**********************************************************************************************************/

class Solution{
public:
    string minWindow(string S, string T){
        if(S.empty()) return "";
        if(S.size() < T.size()) return "";
        const int ASCII_MAX = 256;
        
        int appeared_count[ASCII_MAX];
        int expected_count[ASCII_MAX];
        fill(appeared_count, appeared_count + ASCII_MAX,0);
        fill(expected_count, expected_count + ASCII_MAX,0);
        for(size_t i = 0;,i < T.size(); i++) expected_count[T[i]]++;
        int minWidth = INT_MAX, min_start = 0;
        int wnd_start = 0;
        int appeared = 0;
        for(size_t wnd_end = 0; wnd_end < S.size(); wnd_end ++){
            if(expected_count[S[wnd_end]] > 0){
                appeared_count[S[wnd_end]]++;
                if(appeared_count[S[wnd_end]] <= expected_count[S[wnd_end]])
                    appeared++;
            }
            if(appeared == T.size()){
                while(appeared_count[S[wnd_start]] > expected_count[S[wnd_start]] ||
                   expected_count[S[wnd_start]] == 0 ){
                    appeared_count[S[wnd_start]]--;
                    wnd_start++;
                }
                if(minWidth > (wnd_end - wnd_start + 1)){
                    minWidth = wnd_end - wnd_start + 1;
                    min_start = wnd_start;
                }
            }
        }
        if(minWidth == INT_MAX ) return "";
        else return S.substr(min_start, minWidth);      
    }
};