[LeetCode] Minimum Window Substring

最新推荐文章于 2017-09-07 21:24:22 发布

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最新推荐文章于 2017-09-07 21:24:22 发布

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本文介绍了一种在字符串S中寻找包含字符串T所有字符的最短子串的高效算法，复杂度为O(n)。通过具体示例展示了如何实现该算法，并提供了完整的C++代码实现。

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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.

class Solution {
public:
    string minWindow(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        int ns=S.length();
	    int nt=T.length();
	    vector<int> need(256,0);
	    vector<int> had(256,0);
	    for(int i=0;i<nt;i++)
	    {
		    need[ T[i] ]++;
		    had[ T[i] ]++;
	    }
	    int minLen=ns+100;
	    int mStart=-1;
	    int pre=0,nCount=nt;
	    for(int i=0;i<ns;i++)
	    {
		    if (need[ S[i] ]> 0)
		    {
			    had[ S[i] ]--;
			    if ( had[S[i]]>=0 )
				    nCount--;
		    }
		    if ( nCount ==0 )
		    {
			    while( need[S[pre]]==0 
					    || had[S[pre]]<0 )
			    {
				    if ( need[S[pre]]>0 )
					    had[S[pre]]++;
				    pre++;
			    }
			    if ( i-pre+1 <minLen )
			    {
				    minLen=i-pre+1;
				    mStart=pre;
			    }
		    }
	    }
	    if ( minLen==ns+100 )
		    return "";
	    else
		    return S.substr(mStart,minLen);
    }
};