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:
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; } };
Github 同步地址:
https://github.com/grandyang/leetcode/issues/205
类似题目:
参考资料:
https://leetcode.com/problems/isomorphic-strings/
https://leetcode.com/problems/isomorphic-strings/discuss/57796/My-6-lines-solution
https://leetcode.com/problems/isomorphic-strings/discuss/57810/Short-Java-solution-without-maps