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.

For example,
Given “egg”, “add”, return true.

Given “foo”, “bar”, return false.

Given “paper”, “title”, return true.

Note:
You may assume both s and t have the same length.

先说一下：
据一对一映射的特点，我们需要用两个哈希表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128个字符，所以我们可以用一个128大小的数组来代替哈希表，并初始化为0，我们遍历原字符串，分别从源字符串和目标字符串取出一个字符，然后分别在两个哈希表中查找其值，若不相等，则返回false，若相等，将其值更新为i + 1

代码如下;

int m1[] = new int[128], m2[] = new int[128], n = s.length();        for (int i = 0; i < n; ++i) {            //比如aba            //比如cfc            //i=0    m1[97]=0   m2[99]=0            //i=1    m1[98]=0   m2[102]=0            //Now WILL APPERAR the number had apperaed。            //i=2    m1[97]=1   m2[99]=1 ------> 从这里看出同构 如果不同构的话，那么m2[]此刻的值肯定不等于1            if (m1[s.charAt(i)] != m2[t.charAt(i)])                 return false;            //下面的作用相当于键 起标识作用            //现在s.charAt(i)对应的ASCII码对应的位置等于i+1;            //m1[97] = 1            //m1[98]= 2            m1[s.charAt(i)] = i + 1;            //现在t.charAt(i)对应的ASCII码对应的位置等于i+1;            //m2[99] = 1;            //m2[102] = 2;            m2[t.charAt(i)] = i + 1;        }        return true;