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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：题目中T可能会有重复的字母 我用了各种分类讨论 结果超时
直接看大神的代码吧
class Solution { public: string minWindow(string S, string T) { int m = S.size(), n = T.size(); if (n <= 0 || m < n) return ""; int require[128] = {0}; //记录每种字母需要多少个 关键点 for (int i = 0; i < n; ++i) require[T[i]]++; int count = 0; int minLen = INT_MAX, minIndex = 0; for (int s = 0, e = 0; e < m; ++e) { require[S[e]]--; //末尾数字的需求量减1 if (require[S[e]] >= 0) count++; //如果需求量大于等于0 说明匹配上了新的数字 while (count == n) { //所有字母都被匹配上了 if (e - s + 1 < minLen) { //长度变小了 记录下新的长度 和 起始位置 minLen = e - s + 1; minIndex = s; } require[S[s]]++; //起始位置向后移更新需求量 if (require[S[s]] > 0) count--; s++; } } if (minLen == INT_MAX) return ""; return S.substr(minIndex, minLen); } };
【leetcode】Minimum Window Substring (hard) ★
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原文地址：http://www.cnblogs.com/dplearning/p/4328666.html
