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.

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

还是用一维动态规划中的“局部最优和全局最优法”。

内循环怎么写: dp[i] = U^w _{j = 1,} ( w = i^1/2) min(dp[i], dp[i - j*j] + 1);

写完之后做个test case, 看看初始值怎么写. 

Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;

public int numSquares(int n) {
        int[] dp = new int[n+1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;
        for (int i=1; i<=n; i++) {
            int sqrt = (int)Math.sqrt(i);
            for (int j=1; j<=sqrt; j++) {
                dp[i] = Math.min(dp[i], dp[i-j*j]+1);
            }
        }
        return dp[n];
    }