Integer splitting [dynamic programming]

343. Integer Splitting
     Given a positive integer n, split it into the sum of k positive integers (k >= 2) and maximize the product of these integers.
     Returns the largest product you can get.
     

class Solution {
    
    
    public int integerBreak(int n) {
    
    
        int[] dp = new int[n + 1];//正整数，根据dp数组的定义，从1到n，加上0位置，共n + 1个位置

        dp[2] = 1; 

        for (int i = 3; i <= n; i++) {
    
    
            for (int j = 0; j <= i - j; j++) {
    
    
                dp[i] = Math.max(dp[i], Math.max(j * (i - j), j * dp[i - j]));//累计比较拆分为2个，多个数的最大乘积
            }
        }
        return dp[n];
    }
}