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出处 LeetCode 算法第205题 给定两个字符串 s 和 t,判断它们是否是同构的。 如果 s 中的字符可以被替换得到 t ,那么这两个字符串是同构的。 所有出现的字符都必须用另一个字符替换,同时保留字符的顺序。两个字符不能映射到同一个字符上,但字符可以映射自己本身。 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 说明: 你可以假设 s 和 t 具有相同的长度。
出处 LeetCode 算法第205题
给定两个字符串 s 和 t,判断它们是否是同构的。
如果 s 中的字符可以被替换得到 t ,那么这两个字符串是同构的。
所有出现的字符都必须用另一个字符替换,同时保留字符的顺序。两个字符不能映射到同一个字符上,但字符可以映射自己本身。
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 说明: 你可以假设 s 和 t 具有相同的长度。
用map来保存出现过的映射
/** * @param {string} s * @param {string} t * @return {boolean} */ function helper(s, t) { var map = {}; var sLength = s.length; var tLength = t.length; if (sLength != tLength) return false; var flag = true; for (var i = 0; i < sLength; i++) { var current = s[i]; var target = t[i]; if (current in map) { if (map[current] == target) { continue; } else { flag = false; break; } } else { map[current] = target; } } return flag; } var isIsomorphic = function (s, t) { return helper(s, t) && helper(t, s); };
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习题
思路
用map来保存出现过的映射
解答
The text was updated successfully, but these errors were encountered: