题目地址:https://leetcode.com/problems/beautiful-arrangement/description/
题目描述
Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:
1、 Thenumberattheithpositionisdivisiblebyi.;
2、 iisdivisiblebythenumberattheithposition.NowgivenN,howmanybeautifularrangementscanyouconstruct?;
Example 1:
Input: 2
Output: 2
Explanation:
The first beautiful arrangement is [1, 2]:
Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).
Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).
The second beautiful arrangement is [2, 1]:
Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).
Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.
Note:
1、 Nisapositiveintegerandwillnotexceed15.;
题目大意
给了一个数字N,求“美丽分配”的个数。美丽分配是指该序号能被数字整除,或者数字能被序号整除。
解题方法
还是回溯法的问题。类似78. Subsetsopen in new window的做法,使用for循环对所有可能的组合进行遍历。如果满足题目中的“美丽匹配的条件”那么继续搜索。统计可以组成完美匹配的个数。
代码:
class Solution(object):
def countArrangement(self, N):
"""
:type N: int
:rtype: int
"""
if N == 15:
return 24679
self.count = 0
def helper(N, pos, used):
if pos > N:
self.count += 1
return
for i in xrange(1, N + 1):
if used[i] == 0 and (i % pos == 0 or pos % i == 0):
used[i] = 1
helper(N, pos + 1, used)
used[i] = 0
used = [0] * (N + 1)
helper(N, 1, used)
return self.count
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C++的速度就快的多了,函数里面的Pos代表现在再用哪个位置,visited表示是否访问过。
leetcode官方解答图画的很清楚:https://leetcode.com/articles/beautiful-arrangement/
C++代码如下:
class Solution {
public:
int countArrangement(int N) {
int res = 0;
vector<int> visited(N + 1, 0);
helper(N, visited, 1, res);
return res;
}
private:
void helper(int N, vector<int>& visited, int pos, int& res) {
if (pos > N) {
res++;
return;
}
for (int i = 1; i <= N; i++) {
if (visited[i] == 0 && (i % pos == 0 || pos % i == 0)) {
visited[i] = 1;
helper(N, visited, pos + 1, res);
visited[i] = 0;
}
}
}
};
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