repeated-substring-pattern

https://leetcode.com/problems/repeated-substring-pattern/

下面这个方法，开始我觉得挺好。可惜还是超时了。后来我就加了一个剪枝策略，只有长度能够整除总长度的子串，才需要进行比较。

package com.company;

import java.util.*;

class Solution {
    public boolean repeatedSubstringPattern(String str) {

        for (int i=; i<=str.length()/; i++) {
            if (str.length() % i != ) {
                continue;
            }
            StringBuilder sb = new StringBuilder();
            sb.append(str.substring(i));
            sb.append(str.substring(, i));
            if (str.equals(sb.toString())) {
                return true;
            }
        }
        return false;
    }
}

public class Main {

    public static void main(String[] args) throws InterruptedException {

        System.out.println("Hello!");
        Solution solution = new Solution();

        // Your Codec object will be instantiated and called as such:
        String str = "abcabcabcc";
        boolean ret = solution.repeatedSubstringPattern(str);
        System.out.printf("ret:%b\n", ret);

        System.out.println();

    }

}

下面是开始超时的方法，少了一个剪枝条件。

package com.company;

import java.util.*;

class Solution {
    public boolean repeatedSubstringPattern(String str) {

        for (int i=; i<=str.length()/; i++) {
            StringBuilder sb = new StringBuilder();
            sb.append(str.substring(i));
            sb.append(str.substring(, i));
            if (str.equals(sb.toString())) {
                return true;
            }
        }
        return false;
    }
}

public class Main {

    public static void main(String[] args) throws InterruptedException {

        System.out.println("Hello!");
        Solution solution = new Solution();

        // Your Codec object will be instantiated and called as such:
        String str = "abcabcabcc";
        boolean ret = solution.repeatedSubstringPattern(str);
        System.out.printf("ret:%b\n", ret);

        System.out.println();

    }

}