(M)Backtracking:526. Beautiful Arrangement
2017-10-18 22:21
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这道题的意思是,给一个数字N,从1到N的这些书构成一个序列,使得序列中第i个数满足:i能整除这个数,或者这个数能整除i。问给定的N,满足这个要求的序列能有多少个。想到了卖火柴的小女孩那个题,Matchsticks to Square。这个题里出现一个问题是,求一个数组能不能分成四个和相等的数组。写法和这个差不多。遍历每一个位置i,看i位置能放哪个数,放下这个数后对i+1递归,然后i这个位置试着放下一个符合要求的数,再在这种情况下递归……
class Solution {
public:
int countArrangement(int N) {
int res = 0;
vector<int> visit(N, 0);
back(res, 1, visit);
return res;
}
void back(int &res, int i, vector<int>& visit)
{
if(i == visit.size() + 1){
res++;
return;
}
for(int j = 1; j <= visit.size(); ++j)
{
if(visit[j-1] == 0 && (j % i == 0 || i % j == 0))
{
visit[j-1] = 1;
back(res, i+1, visit);
visit[j-1] = 0;
}
}
}
};
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