Given a positive integer n, find the least number of perfect square numbers (for example,
1, 4, 9, 16, ...
) which sum to n.
For example, given n =
12
, return 3
because 12 = 4 + 4 + 4
; given n = 13
, return 2
because 13 = 4 + 9
.[Thoughts]
一个一维的DP,比较简单。具体看code
[Code]
1: int numSquares(int n) {
2:
3: vector<int> dp(n+1, 0);
4: for(int i =1; i<= n; i++) {
5:
6: int min_v = INT_MAX;
7: for(int j = 1; j*j <= i; j++) {
8: min_v = min(min_v, dp[i - j*j] +1);
9: }
10: dp[i] =min_v;
11: }
12:
13: return dp[n];
14: }
No comments:
Post a Comment