686. Repeated String Match 重复字符串匹配

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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.

给定两个字符串A和B，找到必须重复的最小次数，使得B是其子串。如果没有这样的解决方案，返回-1。

/**
 * @param {string} A
 * @param {string} B
 * @return {number}
 */
var repeatedStringMatch = function (A, B) {
    if (A.includes(B)) {
        return 1;
    }
    let time = 1;
    let isLonger = false;
    while (isLonger == false) {
        let res = A.repeat(++time);
        if (res.includes(B)) {
            return time;
        }
        if (res.length > B.length) {
            isLonger = true;
        }
    }
    return -1;
};

686. Repeated String Match 重复字符串匹配

标签：imu   includes   ica   ted   robot   col   pretty   long   min   

原文地址：http://www.cnblogs.com/xiejunzhao/p/7627998.html