和第一种解法思路类似,依旧是借助字符串的特性,使用替换来验证最大公约数,通过String的replaceAll方法实现。 publicStringgcdOfStrings2(String str1, String str2){if(!(str1+str2).equals(str2+str1)) {return""; }intn=Math.min(str1.length(), str2.length());for(inti=n; i>=1; i--) {if(...
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”。 返回最长字符串 X,要求满足 X 能除尽 st...
(Easy)1071. Greatest Common Divisor of Strings- LeetCode 2019-09-19 18:06 −... CodingYM 0 465 LeetCode 1123. Lowest Common Ancestor of Deepest Leaves 2019-12-15 10:58 −原题链接在这里:https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/ 题目: Given a rooted bin...
The space complexity is O(1) constant, and the time complexity is O(M/N) where M is the length of the longer string and N is the shorter length e.g. “A”, and “AAAAAAA…” See also:Teaching Kids Programming – Greatest Common Divisor of Strings ...
Greatest Greatest Common Divisor Problem Description Andrew has just made a breakthrough in sociology...And the quality of friendship between two persons is equ...
Divide and Conquer Binary Search Math Greatest Common Divisor Prime Knapsack Probability Shuffle Bitmap Basics Misc Bit Manipulation Part II - Coding String Implement strStr Two Strings Are Anagrams Compare Strings Group Anagrams Longest Common Substring Rotate String Reverse Words ...
// find longer of two strings const string &shorterString(const string &s1, const string &s2) { return s1.size() < s2.size() ? s1 : s2; } 1. 2. 3. 4. 5. 形参和返回类型都是指向 const string 对象的引用,调用函数和返回结果时,都没有复制这些 string ...
Many of Ramanujan's results are so inspirational that there is a periodical dedicated to them. The theories of strings and crystals have benefited from Ramanujan's work. (Today some professors achieve fame just by finding a new proof for one of Ramanujan's many results.) Unlike Abel, who in...
Greatest Greatest Common Divisor Problem Description Andrew has just made a breakthrough in sociology...And the quality of friendship between two persons is equ...
lc1071 Greatest Common Divisor of Strings 找两个字符串的最长公共子串 假设:str1.length > str2.length 因为是公共子串,所以str2一定可以和str1前面一部分匹配上,否则不存在公共子串。 所以我们比较str2和str1的0~str2.length()-1部分, 若不同,则直接返回””,不存在公共子串。