Can you solve this real interview question? Greatest Common Divisor Traversal - You are given a 0-indexed integer array nums, and you are allowed to traverse between its indices. You can traverse between index i and index j, i != j, if and only if gcd(nu
Can you solve this real interview question? Greatest Common Divisor of Strings - For two strings s and t, we say "t divides s" if and only if s = t + t + t + ... + t + t (i.e., t is concatenated with itself one or more times). Given two strings str1 and
Github 同步地址: https://github.com/grandyang/leetcode/issues/1071 参考资料: https://leetcode.com/problems/greatest-common-divisor-of-strings/ https://leetcode.com/problems/greatest-common-divisor-of-strings/discuss/307242/C%2B%2B-3-lines https://leetcode.com/problems/greatest-common-divisor-o...
例如示例中的str1 ="ABCABC",str2 ="ABC",str1拼接str2后变成"ABCABCABC",str2拼接str1后变成"ABCABCABC"。而str1 ="LEET",str2 ="CODE",str1拼接str2后变成"LEETCODE",str2拼接str1后变成"CODELEET",两者显然不相等,肯定不存在公约数。 那怎么找到他们的最大公约数呢? 思路:借助字符串拆分。用不...
Greatest Common Divisor of Strings (Java版; Easy) welcome to my blog LeetCode 1071. Greatest Common Divisor of Strings (Java版; Easy) 题目描述 第一次做; 暴力; 核心:1)如果两个字符串str1和str2的最大公因字符串是s, 设该字符串长度为n, 那么str1.substring(0,n)一定等于s, str2.substring...
[leetcode] 1071. Greatest Common Divisor of Strings Description For two strings s and t, we say “t divides s” if and only if s = t + … + t (t concatenated with itself 1 or more times) Given two strings str1 and str2, return the largest string x such that x divides both ...
LeetCode | 1071. Greatest Common Divisor of Strings字符串的最大公因子【Python】 LeetCode 1071. Greatest Common Divisor of Strings字符串的最大公因子【Easy】【Python】【字符串】 Problem LeetCode For strings S and T, we say "T divides S" if and only if S = T + ... + T (T ...
LeetCode 1071. Greatest Common Divisor of Strings字符串的最大公因子【Easy】【Python】【字符串】 Problem LeetCode For stringsSandT, we say "TdividesS" if and only ifS = T + ... + T(Tconcatenated with itself 1 or more times)
1071. Greatest Common Divisor of Strings(字符串的最大公因子) 链接 https://leetcode-cn.com/problems/greatest-common-divisor-of-strings 题目 对于字符串 S 和 T,只有在 S = T + ... + T(T 与自身连接 1 次或多次)时,我们才认定 “T 能除尽 S”。
welcome to my blog LeetCode 1071. Greatest Common Divisor of Strings (Java版; Easy) 题目描述 第一次做; 暴力; 核心:1)如果两个字符串str1和str2的最大公因字符串是s, 设该字符串长度为n, 那么str1.substring(0,n)一定等于s, str2.substring(0,n)一定等于s 第一次做;...【...