leetcode459

1、 Repeated Substring Pattern
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.)

class Solution {
public:
    bool repeatedSubstringPattern(string str) {
        int n=str.size();
        vector<int> dp(n+1,0);
        int i=1,j=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;
    }
};

遍历字符间对比~str[i],str[j]
有几下几种情况：
一、str[i]==str[j]：1、相同字母为小组内；2、相同字母为下一组同位置的字母。
操作：++i，++j，将相同字母的序号存入容器dp[i]=j;
2、str[i]!=str[j]： 1、当j为0时，在之前还没遇到相同字母对，直接 ++j即可；2、当j不为0时，之前已经遇到相同字母对，如果当前的dp[j]已经被存入，则回到上次与str[i]相同时的位置，否则，回到0；