Leetcode205 Isomorphic String

1. 题意理解：

给出两条string，让判断它们是否是同构（isomorphic). 

Isomorphic: two strings are isomorphic if the characters in string s can be replaced to get string t.

2. 解题关键词：hashmap

   1)首先建立一个hashmap，key 对应的是来自s的char，value 对应的是来自t 的 char，以此来记录分别来自string s 和 string t 中每个char的映射。

     比如 "paper" 和 “title”：

   key | val

     p -> t    p as a key does not exists in map, check if t as value has existed in map, if yes, return false; if not, add p as key and t as value.

     a -> i    a as a key does not exists in map, check if i as value has existed in map, if yes, return false; if not, add p as key and i as value.

     p -> t    p as a key already existed in map. check if its corresponding value equals t, if yes, skip, if not, return false.

     e -> l    e as a key does not exists in map, check if l as value has existed in map, if yes, return false; if not add e as key and l as value.

     r -> e   r as a key does not exists in map,  check if e as value has existed in map, if yes, return fals; if not add r as key and e as value.

    DONE.

3. 注意事项：多练几遍。。

4. Java 代 class Solution {

public boolean isIsomorphic (String s, String t) {
        Map<Character, Character> charStoCharT = new hashMap<>();
        int sLength = s.length();
        int tLength = t.length();
       
         //base case: 如果两条string长度根本不同，
         //已经没有必要再比下去，对比的最基本条件是当两条string的长度相同。
        if (sLength != tLength) {
           return false;
        }

        for (int i = 0; i < sLength; i++) {
            char fromS = s.charAt(i);
            char fromT = t.charAt(i);
            if (! charStoCharT.containsKey(fromS)) {
               //可能1: 如果来自s的char不在map里，首先要确定与之对应的来自t的char是否已经 
               //在map里。如果已经在，直接返回false。如果不在，建立新的映射。
               
               if (charStoCharT.containsValue(fromT)) {
                  return false;
               } else {
                  charStoCharT.put(fromS, fromT);
               }
             } else { 
              //可能2:来自s的char已经在map，这时就要把它对应的value拎出来看看是不是现在当前的from t， 如果是，略过，如果不是，返回false.
               if (charStoCharT.get(fromS) != from T) return false;
               else {
               continue;
               }
          }
          return true; //顺利走完return true

}