**Leetcode 76. Minimum Window Substring

最新推荐文章于 2025-09-05 23:16:08 发布

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最新推荐文章于 2025-09-05 23:16:08 发布

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分类专栏： leetcode 笔记文章标签： leetcode

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本文详细解析了如何寻找字符串S中能覆盖字符串T所有字符的最短子串。通过两个示例性的C++实现方案，展示了高效算法的设计思路与实现细节。第一种方法采用多种数据结构辅助查找；第二种方法则简化了过程，使用固定大小的映射表来减少查找次数。

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.

class Solution {
public:
    string minWindow(string s, string t) {

        unordered_set<char> st;

        unordered_map<char, list<size_t> > mp;
        unordered_map<char, size_t> num;
        unordered_map<char, size_t> check;

        vector <size_t> next(s.size());
        char nextfind;
        size_t begin = 0;
        size_t last = begin;
        int flag = 0;

        for (auto ch : t) {
            ++num[ch];
        }

        check = num;
        string ret;
        size_t minlen = s.size() + 1;
        list<int> ls; 
        size_t i = 0; 
        for (; i <= s.size(); ++i) {
            char ch = s[i];
            if (num.end() != num.find(ch)) {

                begin = i;
                last = begin;
                mp[ch].push_back(i);
                if (check.end() != check.find(ch)) {
                    --check[ch];

                }
                if (check[ch] == 0) 
                    check.erase(ch);
                break;
            }
        }
        if (t.size() == 1 && s[begin] == t[0])
            return t;
        for (++i; i <= s.size(); ++i) {
            char ch = s[i];
            if (num.end() != num.find(ch)){
                next[last] = i;
                last = i;

                if (ch == s[begin] && num[ch] == mp[ch].size()) {
                    mp[ch].erase(mp[ch].begin());
                    mp[ch].push_back(i);
                    begin = next[begin];

                    while (mp[s[begin]].size() > num[s[begin]]) {
                        mp[s[begin]].erase(mp[s[begin]].begin());
                        begin = next[begin];
                    }

                }
                 else {
                     mp[ch].push_back(i);
                     if (check.end() != check.find(ch)) {
                        --check[ch];
                         if (check[ch] == 0) 
                            check.erase(ch);
                     }

                     if ((!flag && check.empty()) || (flag && s[i] == nextfind)) { // 找到了一个窗
                         flag = 1;
                        if (i - begin + 1 < minlen) {
                            minlen = i - begin + 1;
                            ret = string(s.begin() + begin, s.begin() + i + 1);

                          }   
                            // 找下一个begin;
                            nextfind = s[begin];
                            mp[s[begin]].erase(mp[s[begin]].begin());
                            if (mp[s[begin]].size() == 0)
                                    mp.erase(s[begin]);
                            begin = next[begin];
                            while (mp[s[begin]].size() > num[s[begin]]) {
                                mp[s[begin]].erase(mp[s[begin]].begin());
                                begin = next[begin];
                            }             
                    }
                }   
            }
        }
        return ret;
    }
};

參考後
别人的代码不是一般的牛逼

class Solution {
public:

    string minWindow(string s, string t) {

        vector<int> map(128, 0);
        size_t counter = t.size();
        size_t head;
        size_t begin = 0;
        size_t len = s.size() + 1;
        for (auto ch : t) {
            ++map[ch];
        }


        for (size_t i = 0; i != s.size(); ++i) {
            char ch = s[i];
            if (map[ch] > 0)
                --counter;
            --map[ch];
            while (counter == 0) {
                if (i - begin + 1 < len) {
                    len = i - begin + 1;
                    head = begin;
                }
                if (map[s[begin]] == 0) ++counter;

                ++map[s[begin]];
                ++begin;
            }
        }
        if (len == s.size() + 1) return "";
        return s.substr(head, len);
    } 
};