LeetCode - 459. Repeated Substring Pattern - O(n)和O(n^2)两种思路 - KMP - (C++) - 解题报告

题目

题目链接
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.)
给定一个没有空格的的字符串，查看该字符串是否由其中的一个子串重复多次得到。给定的字符串中只有小写的英文字母，而且长度不超过10000.

例子 1:
输入: "abab"
输出: True
解释：这是由子串"ab"重复得到

例子 2:
输入: "aba"
输出: False

例子3 3:
输入: "abcabcabcabc"
输出: True
解释：这是由子串"abc"重复四次得到的，也可以认为是"abcabc重复两次得到的"

代码

1.O(n^2)解法

class Solution {
public:
    bool repeatedSubstringPattern(string str) {
        int m=str.size();
        for(int p=1;p<m/2+1;p++)
            for(int i=0;i<m-p;i++)
            {
                if(str[i]!=str[i+p]) break;
                if(m%p==0&&i==m-p-1) return true;
            }
         return false;
    }
};

2.O(n)解法

解法链接这是其他人post到讨论区的解法，基于KMP思想做的，时间复杂性比较好。KMP算法可以参考这个链接

class Solution {
public:
    bool repeatedSubstringPattern(string str) {
        int i = 1, j = 0, n = str.size();
        vector<int> dp(n+1,0);
        while( i < str.size() ){
            if( str[i] == str[j] ) dp[++i]=++j;
            else if( j == 0 ) i++;
            else j = dp[j];
        }
        return dp[n]&&dp[n]%(n-dp[n])==0;
    }