Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string‘s permutations is the substring of the second string.
Example 1:
Input:s1 = "ab" s2 = "eidbaooo" Output:True Explanation: s2 contains one permutation of s1 ("ba").
Example 2:
Input:s1= "ab" s2 = "eidboaoo" Output: False
Note:
- The input strings only contain lower case letters.
- The length of both given strings is in range [1, 10,000].
维护一下一段区间的字符串就行,有O(26*N)的做法,我这里是用multiset做得,时间复杂度理论上 不必O(26*n)大啊但是跑的慢 可能是判断集合是否相等的时候比较费时间 有时间好好看看multiset源码
class Solution { public: bool checkInclusion(string s1, string s2) { multiset<char>se1,se2; if (s2.size() < s1.size()) return false; for (int i = 0; i < s1.size(); ++i) se1.insert(s1[i]),se2.insert(s2[i]); if (se2 == se1) return true; for (int i = s1.size(); i < s2.size(); ++i) { char c = s2[i - s1.size()]; auto x = se2.find(c); se2.erase(x); se2.insert(s2[i]); if (se1 == se2) return true; } return false; } };
时间: 2024-10-11 17:44:11