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面试题 01.02. 判定是否互为字符重排-辅助数组算法
2022-07-07 14:15:00 【Mr Gao】
面试题 01.02. 判定是否互为字符重排
给定两个字符串 s1 和 s2,请编写一个程序,确定其中一个字符串的字符重新排列后,能否变成另一个字符串。
示例 1:
输入: s1 = “abc”, s2 = “bca”
输出: true
示例 2:
输入: s1 = “abc”, s2 = “bad”
输出: false
对于这一题,其实比较简单,我们添加一个辅助数组去判断问题就可以了,做一个辅助数组,存储s1中各个字母的数量 ,然后,对s2中存储的字母数量进行相减,辅助数组最后存储各个字母数量都为0,返回true,否则返回false
解题代码如下:
bool CheckPermutation(char* s1, char* s2){
int r[26];
int i=0;
for(i=0;i<26;i++){
r[i]=0;
}
i=0;
while(s1[i]!='\0'){
r[s1[i]-'a']++;
i++;
}
i=0;
while(s2[i]!='\0'){
r[s2[i]-'a']--;
i++;
}
for(i=0;i<26;i++){
if(r[i]!=0){
return false;
}
}
return true;
}
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