leetcode:Minimum Window Substring

最新推荐文章于 2025-02-20 13:41:31 发布

martin_liang

最新推荐文章于 2025-02-20 13:41:31 发布

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本文介绍了一种在字符串S中寻找包含字符串T所有字符的最短子串的算法，复杂度为O(n)。通过使用两个哈希映射分别记录目标字符串T与源字符串S中的字符出现次数，实现高效查找。

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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).

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 empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

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class Solution {
public:
    string minWindow(string s, string t) {
        
        map<char, int> T_map;
        
        for (int i=0; i<t.size(); i++)
        {
            if (T_map.find(t[i]) == T_map.end())
            {
                T_map[t[i]] = 1;
            }
            else
            {
                T_map[t[i]]++;
            }
        }
        
        map<char, int> S_map;
        int left = 0;
        int right = 0;
        int matchNum = 0;
        int minLeft = 0;
        int minRight = s.size();
        bool found = false;
        
        // 如何比较两个map的数是否match？
        for (int i=0; i<s.size(); i++)
        {
            if (T_map.find(s[i]) != T_map.end())
            {
                if (S_map.find(s[i]) == S_map.end())
                {
                    S_map[s[i]] = 0;
                }

                if (S_map[s[i]] < T_map[s[i]])
                {
                    matchNum++;
                }
                S_map[s[i]]++;
            }
            
            
            if (matchNum == t.size())
            {
                found = true; //这里也要注意，有可能匹配不上
                right = i;
                while (true) //这个while循环比较关键，是用来找到最小range的
                {
                    if (right-left < minRight-minLeft)
                    {
                        minLeft = left;
                        minRight = right;
                    }
                    
                    int preLeft = left;
                    left++;
                    if (left > right) //这里是要注意的，left 不能大于right
                        break;
                    // 每次通过比较preLeft是否有多余来进行左缩
                    if (S_map.find(s[preLeft]) != S_map.end()) //这里也要注意，如果这个数不在 t里，直接跳过 (abc, bc)
                    {
                        S_map[s[preLeft]]--;
                        if (S_map[s[preLeft]] < T_map[s[preLeft]])
                        {
                            matchNum--;
                            break;
                        }
                    }
                }
            }
        }
        
        if (found)//这里注意
            return s.substr(minLeft, minRight-minLeft+1);
        else
            return "";
    }
};