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 2because 13 = 4 + 9.

思路

这是一道非常典型的动态规划题目。

当前如果为元素n，需要找到这个元素n的平方根，然后依次减去平方根的数目，找到前1,4,9,16个元素的Perfect Squares数目。构成一个DP。

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public class Solution {
    public int numSquares(int n) {
        List<Integer> dp = new ArrayList<Integer>();
        dp.add(0);
        dp.add(1);
        for(int i = 2; i <= n; i++){
            int max_root = (int) Math.floor(Math.sqrt((double) i));
            int min_square = Integer.MAX_VALUE;
            for(int j = 1; j <= max_root; j++){
                int tmp = dp.get(i - j * j) + 1;
                if(tmp < min_square) min_square = tmp;
            }
            dp.add(min_square);
        }
        return dp.get(dp.size()-1);
        
    }
}