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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:
// 解法一 function isSame(s1, s2) { if (s1.length !== s2.length) { return false; } let hashA = {}, hashB = {}; for (let i = 0; i < s1.length; i++) { let a = s1[i], b = s2[i]; if (hashA[a] === hashB[b]) { hashA[a] = hashB[b] = (hashB[b] || 0) + 1; continue; } return false; } return true; } //解法二 function isSame(s1, s2) { if (s1.length !== s2.length) { return false; } let hashA = {}, hashB = {}; for (let i = 0; i < s1.length; i++) { let a = s1[i], b = s2[i]; if (!hashA[a] && !hashB[b]) { hashA[a] = b; hashB[b] = a; } else if (hashA[a] !== b || hashB[b] !== a) { return false; } } return true; }
The text was updated successfully, but these errors were encountered:
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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:
示例 2:
示例 3:
说明:
解答
The text was updated successfully, but these errors were encountered: