343. Integer Break

Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).

Note: You may assume that n is not less than 2 and not larger than 58.

Solution:

Dynamic programming.

The tricky part is for number 1, 2 and 3,which the break product is smaller than itself. 

So any number that break into 2 or 3 should use the number itself instead of dp[2] or dp[3]. (dp[1]=0, dp[2]=1, dp[3] =2)

dp[i] =Math.max (dp[i], dp[j]*dp[i-j]) if j >=4 and j-i >=4.

 public int integerBreak(int n) {  
     int[] dp=new int[n+1];  
     for(int i=2;i<=n;i++){  
       for(int j=1;j<=i/2;j++){  
         dp[i]=Math.max(dp[i],Math.max(j,dp[j])*Math.max(i-j,dp[i-j]));  
       }        
     }  
     return dp[n];  
   }