279. Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

Example 1:

Input: n = 12
Output: 3 
Explanation: 12 = 4 + 4 + 4.

Example 2:

Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.


按完全平方数来分割整数

C++：dp
 1 class Solution {
 2 public:
 3     int numSquares(int n) {
 4         if (n == 0)
 5             return 0 ;
 6         vector<int> squareList = getSquareList(n) ;
 7         vector<int> dp(n+1 , 0) ;
 8         for(int i = 1 ; i <= n ; i++){
 9             int minNum = n;
10             for(int square : squareList){
11                 if(square > i){
12                     break ;
13                 }
14                 minNum = min(minNum,dp[i-square] + 1) ;
15             }
16             dp[i] = minNum ;
17         }
18         return dp[n] ;
19     }
20     
21     vector<int> getSquareList(int n) {
22         vector<int> squareList ;
23         int base = 1 ;
24         int diff = 1 ;
25         while(base <= n){
26             squareList.push_back(base) ;
27             diff += 2 ;
28             base += diff ;
29         }
30         return squareList ;
31     }
32 };

 

 
posted @ 2019-01-16 15:50  __Meng  阅读(123)  评论(0)  编辑  收藏  举报