Leetcode_minimum-window-substring(c++ version)

最新推荐文章于 2025-05-14 17:02:27 发布

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最新推荐文章于 2025-05-14 17:02:27 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/flyupliu/article/details/26348829

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本文介绍了解决LeetCode中的最小窗口子串问题的方法，给出两种实现思路及其对应的C++代码实现。该问题要求在给定的字符串S中找到包含另一字符串T所有字符的最小子串。

地址： http://oj.leetcode.com/problems/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.

思路：这道题比较难, 开始花了比较大的力气只完成了T中没有重复字符的最小窗口(用查找)。后来参考了 https://github.com/soulmachine/leetcode 这里的代码，我已经看到几次在字符串查找中使用开256大小的int数组的方法了，主要是当找到完全包含T的一个窗口后(这个过程有点tricky), 动态维护一个最小窗口. 动态维护的过程比较巧妙, 逻辑上要严谨.

参考代码：

class Solution {
public:
    string minWindow(string S, string T) {
        if(S.empty() || T.empty() || S.length() < T.length())
            return "";
        const int CHAR_SIZE = 256;
        int appeared[CHAR_SIZE], expected[CHAR_SIZE], start = 0, matched = 0, min_start = 0, minlen = INT_MAX;
        memset(appeared, 0, sizeof(appeared));
        memset(expected, 0, sizeof(appeared));
        for(int i = 0; i<T.length(); ++i)
            ++expected[T[i]];
        for(int i = 0; i<S.length(); ++i)
        {
            if(expected[S[i]])
            {
                if(appeared[S[i]]<expected[S[i]])
                {
                    ++matched;
                }
                ++appeared[S[i]];
                
                if(matched == T.length())
                {
                    while(!expected[S[start]] || appeared[S[start]] > expected[S[start]])
                    {
                        if(appeared[S[start]] > expected[S[start]])
                        {
                            --appeared[S[start]];
                        }
                        ++start;
                    }
                    if(minlen > i-start+1)
                    {
                        min_start = start;
                        minlen = i - start + 1;
                    }
                }
            }
        }
        if(matched<T.length())
            return "";
        return S.substr(min_start, minlen);
    }
};

//Second trial, 思路和上一个解法一样

class Solution {
public:
    string minWindow(string S, string T) {
        string ans;
        if(S.empty() || S.length()<T.length())
            return ans;
        vector<int>bucket(256, 0), already(256, 0);
        int tlen = T.length(), head = 0, tail = 0, cnt = 0;
        for(int i = 0; i<tlen; ++i)
            ++bucket[T[i]];
        while(tail<S.length()) {
            if(already[S[tail]] < bucket[S[tail]]) {
                ++already[S[tail]];
                ++cnt;
            } else if(bucket[S[tail]]) {
                ++already[S[tail]];
            } else {
                ++tail;
                continue;
            }
            ++tail;
            if(cnt == tlen) {
                while(head<tail && (!bucket[S[head]] || already[S[head]]>bucket[S[head]])) {
                    if(!bucket[S[head]]){
                        ++head;
                        continue;
                    }
                    --already[S[head++]];
                }
                if(head<tail && (ans.empty() || tail-head<ans.length()))
                    ans = S.substr(head, tail-head);
            }
        }
        return ans;
    }
};