Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.
Example 1:
Input: "abab" Output: True Explanation: It‘s the substring "ab" twice.
Example 2:
Input: "aba" Output: False
Example 3:
Input: "abcabcabcabc" Output: True Explanation: It‘s the substring "abc" four times. (And the substring "abcabc" twice.)
题目大意:
给定一个非空字符串,判断它是否能由它的某个子串重复多次构成。
理 解:
根据数学定义f(x+t) = f(x),遍历找到符合条件的周期t。
代 码 C++:
class Solution { public: bool repeatedSubstringPattern(string s) { // 判断周期t是否存在 int t=1,len,i; len = s.length(); for(;t<=len/2;++t){ if(len%t) continue; // 从第2个周期起均与第一个周期值比较 for(i=t;i<len;++i){ if(s[i]!=s[i%t]) break; } if(i==len) return true; } return false; } };
运行结果:
执行用时 :40 ms, 在所有 C++ 提交中击败了80.82%的用户
内存消耗 :11.6 MB, 在所有 C++ 提交中击败了93.22%的用户
原文地址:https://www.cnblogs.com/lpomeloz/p/11113482.html
时间: 2024-10-11 07:58:37