leetcode 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.

 

思路：dp[i]表示和为i的最少的完全平方数个数。

 1 class Solution {
 2 public:
 3     int numSquares(int n) {
 4         if (n <= 0)
 5             return 0;
 6         vector<int> dp(n + 1, 0);
 7         dp[0] = 0;
 8         for (int i = 1; i <= n; i++) {
 9             dp[i] = INT_MAX;
10             for (int j = 1; j * j <= i; j++) {
11                 dp[i] = min(dp[i], dp[i - j * j] + 1);
12             }
13         }
14         return dp[n];
15     }
16 };

static dynamic programming:

 1 class Solution {
 2 public:
 3     int numSquares(int n) {
 4         if (n <= 0)
 5             return 0;
 6         static vector<int> dp({0});
 7         while (dp.size() <= n) {
 8             int m = dp.size();
 9             int cnt = INT_MAX;
10             for (int j = 1; j * j <= m; j++) {
11                 cnt = min(cnt, dp[m - j * j] + 1);
12             }
13             dp.push_back(cnt);
14         }
15         return dp[n];
16     }
17 };

 

 

思路三：数学方法：勒让德多项式。