343 Integer Break

343. Integer Break

1. Question

Given a positive integern, 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, givenn= 2, return 1 (2 = 1 + 1); givenn= 10, return 36 (10 = 3 + 3 + 4).

Note: You may assume thatnis not less than 2 and not larger than 58.

2. Implementation

(1) DP

思路: dp[i]表示i能分解成的integer后的最大乘积， 对于每个整数i，我们再1 - i的范围内找到一个整数j， 使得dp[i] = Math.max(dp[i], j * Math.max(i - j, dp[i - j]);

class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n + 1];
        dp[1] = 1;

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

3. Time & Space Complexity

DP: 时间复杂度O(n^2), 空间复杂度O(n)