[LeetCode] 205. Isomorphic Strings 同构字符串
Given two strings s and t, determine if they are isomorphic.
Two strings are isomorphic if the characters in s can be replaced to get t.
All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.
Example 1:
Input: s = "egg", t = "add"
Output: true
Example 2:
Input: s = "foo", t = "bar" Output: false
Example 3:
Input: s = "paper", t = "title"
Output: true
Note:
You may assume both s and t have the same length.
这道题让我们求同构字符串,就是说原字符串中的每个字符可由另外一个字符替代,可以被其本身替代,相同的字符一定要被同一个字符替代,且一个字符不能被多个字符替代,即不能出现一对多的映射。根据一对一映射的特点,需要用两个 HashMap 分别来记录原字符串和目标字符串中字符出现情况,由于 ASCII 码只有 256 个字符,所以可以用一个 256 大小的数组来代替 HashMap,并初始化为0,遍历原字符串,分别从源字符串和目标字符串取出一个字符,然后分别在两个数组中查找其值,若不相等,则返回 false,若相等,将其值更新为 i + 1,因为默认的值是0,所以更新值为 i + 1,这样当 i=0 时,则映射为1,如果不加1的话,那么就无法区分是否更新了,代码如下:
1
2
3
4
5
6
7
8
9
10
11
12
|
class Solution {
public :
bool isIsomorphic(string s, string t) {
int m1[256] = {0}, m2[256] = {0}, n = s.size();
for ( int i = 0; i < n; ++i) {
if (m1[s[i]] != m2[t[i]]) return false ;
m1[s[i]] = i + 1;
m2[t[i]] = i + 1;
}
return true ;
}
};
|
到此这篇关于C++实现LeetCode(205.同构字符串)的文章就介绍到这了,更多相关C++实现同构字符串内容请搜索服务器之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持服务器之家!
原文链接:https://www.cnblogs.com/grandyang/p/4465779.html