leetcode - Minimum Window Substring

AkibaTakuya

于 2014-10-26 16:56:02 发布

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本文链接：https://blog.youkuaiyun.com/AkibaTakuya/article/details/40477899

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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)。通过使用计数器和字符集来跟踪匹配情况，实现了高效查找。

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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 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:
    std::string minWindow(std::string S, std::string T) {
        if (S.empty() || T.empty())
        {
            return "";
        }
        int count = T.size();
        int require[128] = {0};
        bool chSet[128] = {false};
        for (int i = 0; i < count; ++i)
        {
            require[T[i]]++;
            chSet[T[i]] = true;
        }
        int i = -1;
        int j = 0;
        int minLen = INT_MAX;
        int minIdx = 0;
        while (i < (int)S.size() && j < (int)S.size())
        {
            if (count)
            {
                i++;
                require[S[i]]--;
                if (chSet[S[i]] && require[S[i]] >= 0)
                {
                    count--;
                }
            }
            else
            {
                if (minLen > i - j + 1)
                {
                    minLen = i - j + 1;
                    minIdx = j;
                }
                require[S[j]]++;
                if (chSet[S[j]] && require[S[j]] > 0)
                {
                    count++;
                }
                j++;
            }
        }
        if (minLen == INT_MAX)
        {
            return "";
        }
        return S.substr(minIdx, minLen);
    }
};