Description

Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1.

For example, with A = “abcd” and B = “cdabcdab”.

Return 3, because by repeating A three times (“abcdabcdabcd”), B is a substring of it; and B is not a substring of A repeated two times (“abcdabcd”).

Note

• The length of A and B will be between 1 and 10000.

Solution

對於String A = "abcd"我們考慮這三個例子：String B = "abcdabcdabcd"、String B = "abcdabcdab"、String B = "cdabcdabcdab"。

在第一種情況中，A和B是完全相同的字串，因此只需要重複字串A三次即可得到B；在第二種情況中，B的前兩個abcd是與A一模一樣的，而第三個只有ab沒有cd，此時依然需要重複字串A三次，B才為A的子字串；對於第三種情況則比較特殊，雖然它也包含了三個abcd

根據以上分析，我們先重複字串A直到它的長度大於或等於字串B，這時候判斷字串B是否為字串A的子字串，如果判斷為否，說明字串B不屬於第一二種情況，則再重複一遍字串判斷其是否為第三種情況，如果是的話則返回遍數，如果不是就說明沒有結果，返回-1。

class Solution {
    public int repeatedStringMatch(String A, String B) {
    	StringBuilder sb = new StringBuilder(A);
    	int
 
 count = 1;
    	while(sb.length() < B.length()) {
    		sb.append(A);
    		count++;
    	}
    	if(sb.toString().contains(B)) return count;
    	if(sb.append(A).toString().contains(B)) return count+1;
    	return -1;
    }
}