LeetCode Minimum Window Substring

最新推荐文章于 2021-02-25 13:50:45 发布

coobj2008

最新推荐文章于 2021-02-25 13:50:45 发布

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本文提供了一种复杂度为O(n)的算法，用于在给定字符串中找到包含指定子字符串的最短子串。通过使用首尾指针，算法能够高效地解决此类问题。

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题目：

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:
    //使用首尾指针，复杂度为O(n)
	string minWindow(string S, string T) {
		if (S.size() == 0 || T.size() == 0)
			return "";
		int n = T.size();
		int minS = INT_MAX;
		int xx;
		int tt1[256] = { 0 };
		int ss[256] = { 0 };
		for (int i = 0; i < n; i++)
			tt1[T[i]]++;
		int end = 0;
		int cnt = 0;
		for (int beg = 0; beg < S.size(); beg++) {
			while (end < S.size()) {
				if (cnt == n)
					break;
				if (tt1[S[end]] > 0) {
					ss[S[end]]++;
					if (ss[S[end]] <= tt1[S[end]]) {
						cnt++;
					}
				}
				if (cnt == n)
					break;
				end++;
			}
			//没有再符合条件的了！！
			if (end == S.size())
				break;
			if (end - beg + 1 < minS) {
				minS = end - beg + 1;
				xx = beg;
			}
			if (tt1[S[beg]] > 0) {
				ss[S[beg]]--;
				if (ss[S[beg]] < tt1[S[beg]])
					cnt--;
			}
			if (cnt != n)
				end++;
		}
		if (minS == INT_MAX)
			return "";
		string ans(S, xx, minS);
			return ans;
	}
};