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.

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

 

 

public int numSquares(int n) {
  int[] dp = new int[n + 1];
  dp[0] = 0;
  for (int i = 1; i <= n; i++) {
    int j = 1;
    int min = Integer.MAX_VALUE;
    while (i - j * j >= 0) {
      min = Math.min(min, dp[i - j * j] + 1);//例如：13包括1，4 ，9三个平方数，dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1两两对比得最小值
      j++;
    }
    dp[i] = min;
  }
  return dp[n];
}