LeetCode Minimum Window Substring

标签：

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：还是双指针发处理，尾指针先找到已经包含了T串的位置，然后头指针再不停的收缩，标记的方式就是记录字符出现的次数

class Solution {
public:
    int count1[256];
    int count2[256];

    string minWindow(string S, string T) {
        if (T.size() == 0 || S.size() == 0)
            return "";
            
        memset(count1, 0, sizeof(count1));
        memset(count2, 0, sizeof(count2));
        
        for(int i = 0; i < T.size(); i++) {
            count1[T[i]]++;
            count2[T[i]] = 1;
        }
        
        int count = T.size();
        
        int start = 0;
        int minSize = INT_MAX;
        int minStart;
        for(int end = 0; end < S.size(); end++) {
            if (count2[S[end]] > 0) {
                count1[S[end]]--;
                if (count1[S[end]] >= 0)
                    count--;
            }
            
            if (count == 0) {
                while(1) {
                    if (count2[S[start]] > 0) {
                        if (count1[S[start]] < 0)
                            count1[S[start]]++;
                        else break;
                    }
                    start++;
                }
            
                if (minSize > end - start + 1) {
                    minSize = end - start + 1;
                    minStart = start;
                }
            }
        }
        
        if (minSize == INT_MAX) return "";
        return S.substr(minStart, minSize);
    }
};

LeetCode Minimum Window Substring

标签：

原文地址：http://blog.csdn.net/u011345136/article/details/43488839