leetcode 466. Count The Repetitions

Define S = [s,n] as the string S which consists of n connected strings s. For example, [“abc”, 3] =”abcabcabc”.

On the other hand, we define that string s1 can be obtained from string s2 if we can remove some characters from s2 such that it becomes s1. For example, “abc” can be obtained from “abdbec” based on our definition, but it can not be obtained from “acbbe”.

You are given two non-empty strings s1 and s2 (each at most 100 characters long) and two integers 0 ≤ n1 ≤ 106 and 1 ≤ n2 ≤ 106. Now consider the strings S1 and S2, where S1=[s1,n1] and S2=[s2,n2]. Find the maximum integer M such that [S2,M] can be obtained from S1.

Example:

Input:
s1=”acb”, n1=4
s2=”ab”, n2=2

Return:
2

计算重复出现的元素，最简单的做法就是暴力搜索，我就喜欢这样的方法，虽然会超时，但是别的方法不太容易做，这道题就这样吧。

多余n个s1组成的S1，假如S1包含n个s2，那么S1就包含n/n2个S2，S2是由n2个s2构成，直接双指针遍历即可

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>using namespace std;class Solution {public:    int getMaxRepetitions(string s1, int n1, string s2, int n2)     {        int i = 0, count1 = 0, j = 0, count2 = 0;        while (count1 < n1)        {            if (s1[i] == s2[j])            {                j++;                if (j == s2.length())                {                    count2++;                    j = 0;                }            }            i++;            if (i == s1.length())            {                count1++;                i = 0;            }        }        return count2 / n2;    }};