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.
1.dp
f(n) = min{f(n-1),f(n-4),f(n-9)...f(n-k)}+1,其中k*k <=n && k>=1;
class Solution { public: int numSquares(int n) { vector<int> dp(n+1,0); for(int i=1;i<=n;i++){ int k = INT_MAX ,j=1; while(j*j <= i){ k = min(k,dp[i-j*j]+1); j++; } dp[i] = k; } return dp.back(); } };
原文：http://www.cnblogs.com/zengzy/p/5058079.html