Day 10 - LeetCode 205. Isomorphic Strings

LeetCode 205. Isomorphic Strings

題目

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.

翻譯

給兩個字串s跟t，判斷他們是否是同構字。
如果他們是同構字，表示s裡面毎個字元都可以拿來對應t的特定字元。
全部的字元都要依順序被取代，而且s一種字元只會對應t一種字元，也可能對應到與自己相同的字元。

備註：
你可以假設 s 跟 t 有相同長度

想法

建立兩個map儲存互相對應, 再比較有沒有被更動過第一次儲存的對應值

詳細參閱

https://discuss.leetcode.com/topic/85389/3ms-java-solution-with-two-char-mappings-beats-99-5

---

Code

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isIsomorphic = function(s, t) {
    //不符合條件先跳出
    if (!s || !t || s.length != t.length) {
        return false;
    }
    //建立map, 輸入字串轉陣列
    var s_map = {};
    var t_map = {};
    var input_s_array = s.split('');
    var input_t_array = t.split('');

    for (let i = 0; i < input_s_array.length; i++) {
        var s_char = input_s_array[i];
        var t_char = input_t_array[i];
        //只要沒有就寫入map
        if (!s_map[s_char] && !t_map[t_char]) {
            s_map[s_char] = t_char;
            t_map[t_char] = s_char;
        }
        //比較是否符合, 跟原本第一次寫入的不一樣就return fale
        if (s_map[s_char] != t_char || t_map[t_char] != s_char) {
            return false;
        }
    }
    return true;
};

Run

Your input
"egg", "add"
"foo", "bar"
"title", "paper"

Your answer
true
false
true

Expected answer
true
false
true

Runtime: 112 ms