[LeetCode]Minimum Window Substring

最新推荐文章于 2017-09-07 21:24:22 发布

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本文介绍了一种寻找字符串S中包含字符串T所有字符的最小子串的高效算法。通过使用滑动窗口技巧，并结合计数表来跟踪所需字符的状态，实现了时间复杂度为O(n)的解决方案。

class Solution {
//need2FT + hasFT + two pointer + count(check if satisfied)
//O(n)
public:
	string minWindow(string S, string T) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int lenT = T.size();
		int lenS = S.size();
		if(lenT <= 0 || lenT > lenS) return string("");//illegal input
		int need2FT[256] = {0};
		int hasFT[256] = {0};
		for (int i = 0; i < lenT; ++i)//init need2FT 
			need2FT[T[i]]++;
		int start, end;//two pointer to sliding window
		start = end = 0;
		int count = 0;
		int minStart, minEnd, minLen;
		minLen = INT_MAX;
		for ( ; end < lenS; ++end)
		{
			if(0 == need2FT[S[end]]) continue;//unrelated character
			hasFT[S[end]]++;
			if (hasFT[S[end]] <= need2FT[S[end]])
				count++;
			if (count == lenT)//maintain the minimum sliding window
			{
				while (hasFT[S[start]] > need2FT[S[start]] || 0 == need2FT[S[start]])
				{
					if(0 != need2FT[S[start]])
						hasFT[S[start]]--;
					start++;
					if(start >= lenS)
						break;
				}
				int newWinLen = end-start+1;
				if (newWinLen < minLen)
				{
					minStart = start; 
					minEnd = end;
					minLen = newWinLen;
				}
			}
			if(start >= lenS) break;
			
		}
		if(minLen == INT_MAX) return string("");
		//then return the qualified sliding window
		return S.substr(minStart, minLen);
	}
};

second time

class Solution {
public:
    string minWindow(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> cntTable(256, 0);
        vector<int> checkTable(256, 0);
        for(int i = 0; i < T.size(); ++i) checkTable[T[i]]++;
        //find the first window
        int validCnt = 0;
        int end;
        for(end = 0; end < S.size(); ++end)
        {
            if(cntTable[S[end]] < checkTable[S[end]]) validCnt++;
            cntTable[S[end]]++;
            if(validCnt == T.size()) break;
        }
        if(end == S.size()) return "";
        int minStart = 0;
        while(minStart < S.size() && cntTable[S[minStart]] > checkTable[S[minStart]]) cntTable[S[minStart++]]--;
        int minEnd = end;
        //try to find a minimum one
        int curStart = minStart;
        for(int i = minEnd+1; i < S.size(); ++i)
        {
            cntTable[S[i]]++;
            int newStart = curStart;
            while(cntTable[S[newStart]] > checkTable[S[newStart]])
            {
                cntTable[S[newStart]]--;
                newStart++;
            }
            curStart = newStart;
            if(i-newStart < minEnd-minStart) minStart = newStart, minEnd = i;
        }
        
        return S.substr(minStart, minEnd-minStart+1);
    }
};