[leetcode] 567. Permutation in String
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567. Permutation in String
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].
题目:
给定两个字符串s1和s2,判断s1的排列是否在s2中。
解题思路
方法1:最简单最暴力的方法其实就是找到s1的所有全排列,然后在s2中查找是否这些全排列字符串在s2中。但是这种方法耗时太大,会导致超时。
方法2:滑动窗口
其实不需要找到s1的全排列,因为我们只需要考虑s2中是否包含s1中同样个数的字符,并且这些字符是连在一起的就行了。因此,我们可以使用一个滑动窗口,在s2上滑动。在这个滑动窗口中的字符及其个数是否刚好等于s1中的字符及其个数,此外滑动窗口保证了这些字符是连在一起的。
bool checkInclusion(string s1, string s2) { int len1 = s1.size(); int len2 = s2.size(); if (len1 > len2) return false; vector<int> maps1(26), maps2(26); for (int i = 0; i < len1; ++i) { maps1[s1[i] - 'a'] ++; maps2[s2[i] - 'a'] ++; } if (maps1 == maps2) return true; for (int i = 0; i + len1 < len2; ++i) { maps2[s2[i] - 'a'] --; maps2[s2[i + len1] - 'a'] ++; if (maps1 == maps2) return true; } return false;}
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