[leetcode]第九周作业

题目

来源： < leetcode > 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.

分析与代码

方法一：

令dp
为n对应的加数最大积。

那么递推方程就是：dp
=max(i*dp[n-i],i*(n-i)) (1<=i<=n-1)。

边界: dp[1]=1 dp[2] =1

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], (Math.max(j,dp[j])) * (Math.max(i - j, dp[i - j])));
}
}
return dp
;
}

方法二：

解法来自网上，纯数学方法

看n包含多少个3,把他们相乘,直到n<=2

class Solution {
public:
int integerBreak(int n) {
if(n < 4) return n-1;
int res = 1;
while(n > 2){//看n包含多少个3,把他们相乘,直到n<=2
res *= 3;
n -= 3;
}
if(n == 0) return res;//n可以整除3，res就是各个3相乘
if(n == 1) return (res / 3 ) * 4;//除3余1，把其中的一个3加1变为4再相乘
if(n == 2) return res * 2;//除3余2,则可直接把2与res相乘
}
};