Perfect Squares

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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.

public int numSquares(int n) {       int[] dp = new int[n + 1];       Arrays.fill(dp, Integer.MAX_VALUE);       dp[0] = 0;       for(int i = 0; i <= n; i++){           for(int j = 1; i + j * j <= n; j++){           /**            * dp[i] + 1 中的1代表j*j这个组合            * e.g:dp[0+2*2] = min(dp[4],dp[0]+1)            * 相当于0的组合方式+2*2的组合方式，得到了4的组合方式            *             * e.g:4            * i=0,j=2            * dp[4] = min(dp[4],dp[0]+1) = min(max,0+1) = 1            * i=3,j=1            * dp[4] = min(dp[4],dp[3]+1) = min(1,3+1) = 1            * dp[3] + 1 中的1相当于1*1            */               dp[i  + j * j] = Math.min(dp[i + j * j], dp[i] + 1);            }       }       return dp[n];    }

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