Given two strings,write a method to decide if one is a permutation of the other.
Example 1:
Input: s1 = "abc", s2 = "bca" Output: true Example 2:
Input: s1 = "abc", s2 = "bad" Output: false Note:
0 <= len(s1) <= 100 0 <= len(s2) <= 100
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/check-permutation-lcci 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
https://hanxu.blog.csdn.net/article/details/106053696
首先对 字符串进行预判 然后再进行 字符出现次数统计,判断数量。
class Solution {
public boolean CheckPermutation(String s1, String s2) {
if(s1.length() != s2.length()||s1.length()==0){
return false;
}
int[] num = new int[300];
for(int i=0;i<s1.length();i++){
num[s1.charAt(i)]++;
num[s2.charAt(i)]--;
}
for(int i=0;i<300;i++){
if(num[i]!=0){
return false;
}
}
return true;
}
}