leetcode[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.

Example 1:

Input: n = 12
Output: 3 
Explanation: 12 = 4 + 4 + 4.

Example 2:

Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.

题目大意：

求解一个数最少能被多少个完全立方数相加得到。

解法：

使用动态规划的方法，dp[i]代表的是i最少能被多少个完全立方数相加得到，找到规律如下：

dp[0] = 0 
dp[1] = dp[0]+1 = 1
dp[2] = dp[1]+1 = 2
dp[3] = dp[2]+1 = 3
dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 } 
      = Min{ dp[3]+1, dp[0]+1 } 
      = 1				
dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 } 
      = Min{ dp[4]+1, dp[1]+1 } 
      = 2
						.
						.
						.
dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 } 
       = Min{ dp[12]+1, dp[9]+1, dp[4]+1 } 
       = 2
						.
						.
						.
dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1

java:

class Solution {
    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 j=1;
            while(i-j*j>=0){
                dp[i]=Math.min(dp[i-j*j]+1,dp[i]);
                j++;
            }
        }
        return dp[n];
    }
}