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[leetcode-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].
思路1:
首先s1全排列,每一个全排列 查看是否是s2的子串。果然...超时,复杂度太高。
bool checkInclusion(string s1, string s2) { if(s1.size()>s2.size())return false; char* s1char = new char(s1.size()+1); s1.copy(s1char,s1.size(),0); char* s1char2= new char(s1.size()+1); s1.copy(s1char2,s1.size(),0); do { if(s2.find(s1char) != string::npos) return true; }while(next_permutation(s1char,s1char+strlen(s1char))); do { if(s2.find(s1char2) != string::npos) return true; }while(prev_permutation(s1char2,s1char2+strlen(s1char2))); return false; }
思路2:
学习了已经AC的大神的代码,发现他们无一例外都是用hash表格来做的。
思路就是:观察字符串s1和s2,如果s1的全排列是s2的子串,则s1的所有字符均出现在s2中,而且这些字符连续。
于是用一个类似于滑动窗口的区间去统计此区间内的字符,区间的长度取为s1的长度。
如果在s2中存在一个区间,里面的字符与s1的字符全部都一样,则说明s1通过全排列后一定能够得到s2中的子串。也即满足题意。
bool checkInclusion2(string s1, string s2) { int m = s1.length(),n = s2.length(); if(m>n)return false; vector<int>maps1(26),maps2(26); for(int i=0;i<m;i++) { maps1[s1[i] - ‘a‘] ++; maps2[s2[i] - ‘a‘] ++; } if(maps1 == maps2) return true; //统计前m个字符是否符合条件 for(int i = 0;i+m<n;i++) { maps2[s2[i] - ‘a‘] --; maps2[s2[i+m] - ‘a‘] ++; if(maps1 == maps2) return true; } return false; }
[leetcode-567-Permutation in String]