[Swift]LeetCode343. 整数拆分 |

发布时间：2020-12-14 05:06:09 所属栏目：百科来源：网络整理

导读：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. Example 1: Input: 2Output: 1 Explanation: 2 = 1 + 1,1 × 1 = 1. Exampl

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.

Example 1:

Input: 2
Output: 1 Explanation: 2 = 1 + 1,1 × 1 = 1.

Example 2:

Input: 10
Output: 36 Explanation: 10 = 3 + 3 + 4,3 ×?3 ×?4 = 36.

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

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。

示例 1:

输入: 2
输出: 1
解释: 2 = 1 + 1,1 × 1 = 1。

示例?2:

输入: 10
输出: 36
解释: 10 = 3 + 3 + 4,3 ×?3 ×?4 = 36。

说明:?你可以假设?n?不小于 2 且不大于 58。

8ms

 1 class Solution {
 2     func integerBreak(_ n: Int) -> Int {
 3         if n < 2 {
 4             return 0
 5         }
 6         
 7         var res = Array(repeating: 1,count: n+1)
 8         
 9         res[1] = 0
10         
11         for i in 2...n {
12             var maxRes = 1
13             for j in 1..<i {
14                 maxRes = max(maxRes,max(res[j],j) * max(i-j,res[i-j]))
15             }
16             res[i] = maxRes
17         }
18         return res[n]
19     }
20 }

8ms

 1 class Solution {
 2     func integerBreak(_ n: Int) -> Int {
 3         
 4         guard n > 3 else {
 5             return [1,1,2][n]
 6         }
 7         
 8         var times3 = n / 3
 9         
10         if n % 3 == 1 {
11             times3 -= 1
12         }
13         
14         let times2 = (n - times3 * 3) / 2
15         
16         return Int(pow(3.0,Double(times3))) * Int(pow(2.0,Double(times2)))
17     }
18 }

16ms

 1 class Solution {
 2     func integerBreak(_ n: Int) -> Int {
 3         if n == 2 {
 4             return 1
 5         } else if n == 3 {
 6             return 2
 7         } else if n % 3 == 0 {
 8             return Int(pow(3,Double(n / 3)))
 9         } else if n % 3 == 1 {
10             return Int(2 * 2 * pow(3,Double((n - 4) / 3)))
11         } else { // 2
12             return Int(2 * pow(3,Double((n - 2) / 3)))
13         }
14     }
15 }

24ms

 1 class Solution {
 2     func integerBreak(_ n: Int) -> Int {
 3         var dps = Array(repeating: 0,count: n + 1)
 4         dps[1] = 1
 5         for num in 2...n {
 6             for j in 1..<num {
 7                 dps[num] = max(dps[num],j * max(num - j,dps[num - j]))
 8             }
 9         }
10 
11         return dps[n]
12     }
13 }

（编辑：李大同）

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