Perfect Squares

原来是dp问题，最好解。

转移方程：

dp[i] = Math.min(dp[i], dp[i - j*j] + 1);

求的是最小值

public class Solution {    public int numSquares(int n) {        int max = (int)Math.sqrt(n);int[] dp = new int[n + 1];Arrays.fill(dp, Integer.MAX_VALUE);for (int i = 1; i <= n; i++) {for (int j = 1; j <= max; j++) {if (i == j*j) {dp[i] = 1;} else if(i > j*j) {dp[i] = Math.min(dp[i], dp[i - j*j] + 1);}}}return dp[n];    }}

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.