0730

给定一个正整数 n，将其拆分为至少两个正整数的和，并使这些整数的乘积最大化。 返回你可以获得的最大乘积。

 

动态规划

================================Python=================================

class Solution:
    def integerBreak(self, n: int) -> int:
        dp = [0 for i in range(n+1)]
        for i in range(1, n+1):
            for j in range(i):
                dp[i] = max(dp[i], dp[i-j] * j, j * (i - j))
        return dp[-1]

================================Java==================================

class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n+1];
        for (int i = 2; i <= n; i ++){
            int curMax = 0;
            for (int j = 1; j <= i - 1; j++){
                curMax = Math.max(curMax, Math.max(dp[i - j] * j, (i - j) * j));
            }
            dp[i] = curMax;
        }
        return dp[n];
    }
}

class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n+1];
        for (int i = 1; i <= n; i ++){
            for (int j = 0; j <= i - 1; j++){
                dp[i] = Math.max(dp[i], Math.max(dp[i - j] * j, (i - j) * j));
            }
        }
        return dp[n];
    }
}

=============================Go=============================

func integerBreak(n int) int {
    dp := make([]int, n+1)
    for i := 2; i < n + 1; i++ {
        for j := 1; j < i; j++ {
            dp[i] = max(dp[i], max(dp[i - j] * j, (i - j) * j))
        }
    }
    return dp[n]
}

func max(m, n int) int {
    if (m > n) {
        return m
    } else {
        return n
    }
}