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Day 20. Repeated String Match(68

Day 20. Repeated String Match(68

作者: 前端伊始 | 来源:发表于2018-01-02 23:04 被阅读0次

    问题描述
    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.

    **思路:知道js中的indexOf()方法就够了(返回某个指定的字符串值在字符串中首次出现的位置),扩展lastIndexOf()以及 String直接赋字符串和new String的区别

    **

    *
    /**
     * @param {string} A
     * @param {string} B
     * @return {number}
     */
    var repeatedStringMatch = function(A, B) {
        var n = B.length/A.length;
        var str = "";
        for(var i = 1; i<n+5; i++){
            str += A;
            if(str.indexOf(B)>=0){
               return i;
            }  
        }
        return -1;
    };
    

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