leetcode-isomorphic string judgment

205. 同构字符串
     给定两个字符串 s 和 t ,判断它们是否是同构的.

如果 s 中的字符可以按某种映射关系替换得到 t ,那么这两个字符串是同构的.

每个出现的字符都应当映射到另一个字符,同时不改变字符的顺序.不同字符不能映射到同一个字符上,相同字符只能映射到同一个字符上,字符可以映射到自己本身.

示例 1:

输入：s = “egg”, t = “add”
输出：true
示例 2：

输入：s = “foo”, t = “bar”
输出：false
示例 3：

输入：s = “paper”, t = “title”
输出：true

提示：

1 <= s.length <= 5 * 104
t.length == s.length
s 和 t 由任意有效的 ASCII 字符组成

let isIsomorphic = function(s, t) {
    
	//return different lengthfalse
  if (s.length !== t.length) return false
  let map1 = {
    }//建立映射表
  for (let i = 0; i < s.length; i++) {
    
    if (!map1[s[i]]) {
      //If not in the mapping table, add it
      map1[s[i]] = t[i]
    } else {
      //If there is existing in the mapping table, it is judged whether it conforms to the previous mapping 不符合直接返回false
      if (t[i] !== map1[s[i]]) return false
    }
  }
  //Finally, determine the mapping tablevalueAre all different characters,That is, it meets the requirement that different characters cannot be mapped to the same character Return if notfalse
  if (Array.from(new Set(Object.values(map1))).length !== Object.values(map1).length) return false
  return true
};

console.log(isIsomorphic('ab', 'cc'));