Minimum Window Substring

最新推荐文章于 2024-01-06 14:12:46 发布

brucehb

最新推荐文章于 2024-01-06 14:12:46 发布

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本文提供了一种复杂度为O(n)的算法，用于在一个字符串中找到包含另一个字符串所有字符的最短子串。通过实例演示了如何实现并优化此过程。

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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:
    string minWindow(string s, string t) {
        int m = s.length();
        int n = t.length();

        string result = "";
        if (m < n)
        {
            return result;
        }

        int num[256];
        int count[256];
        memset(num, 0, 256*sizeof(int));
        memset(count, 0, 256*sizeof(int));
        for (int i = 0; i < n; i++)
        {
            num[t[i]]++;
        }

        int left = 0;
        int right = 0;
        int foundChars = 0;
        int minLen = m+1;
        while (right < m)
        {
            int temp = (int)s[right];
            if (num[temp] != 0)
            {
                count[temp]++;
                if (count[temp] <= num[temp])
                {   
                    foundChars++;
                }
            }

            if (foundChars == n)
            {
                while (num[s[left]] == 0 || count[s[left]] > num[s[left]])
                {
                    if (count[s[left]] > num[s[left]])
                    {
                        count[s[left]]--;
                    }
                    left++;
                }
                if (right-left+1 < minLen)
                {
                    minLen = right-left+1;
                    result = s.substr(left, minLen);
                }
                count[s[left]]--;
                foundChars--;
                left++;
            }

            right++;
        }

        return result;
    }
};