76. 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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string "".
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

最小窗口子串，已知字符串S和字符串T,求在S中的最小窗口（区间），使得这个区间中包含字符串T中的所有字符。

解法：采用哈希表记录出现的字符，用双指针界定窗口，寻找包含字符串T中的所有字符的最小窗口。

class Solution {
private:
bool isContain(vector<int> map_cur_s, vector<int> map_t)
{
	int len = map_cur_s.size();
	for (int i = 0; i < len; i++)
	{
		if (map_cur_s[i] < map_t[i])
			return false;
	}
	return true;
}
public:
   string minWindow(string s, string t) {
	if (s.size() < t.size())
		return "";
		
	vector<int> map_t(128,0);
	set<char> log_t;
	for (int i = 0; i < t.size(); i++)
	{
		map_t[t[i]]++;
		log_t.insert(t[i]);
	}
	int begin = 0, i = 0;
	string result;
	int minLen = INT_MAX;
	vector<int> map_cur_s(128, 0);
	while (i <= s.size())
	{
		map_cur_s[s[i]]++;	
		while (begin <= i)
		{
			if (log_t.find(s[begin]) == log_t.end() || map_cur_s[s[begin]] > map_t[s[begin]])
			{
				map_cur_s[s[begin]]--;
				begin++;
			}
			else
			{
				break;
			}
		}

		int count = i - begin + 1;
		if (isContain(map_cur_s, map_t))
		{
			if (count < minLen)
			{
				minLen = count;
				result = s.substr(begin, count);
			}			
		}		
		i++;		
	}
	return result;
}
};