題目
Given an integer array, find three numbers whose product is maximum and output the maximum product.
Example 1:
Input: [1,2,3]
Output: 6
Example 2:
Input: [1,2,3,4]
Output: 24
Note:
- The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
- Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer.
解題思路
主要問題在負(fù)數(shù)的考慮，但是一共三個(gè)數(shù)，所以如果有負(fù)數(shù)的話一定是兩個(gè)負(fù)數(shù)，一個(gè)正數(shù)，這樣的話肯定是最小的兩個(gè)負(fù)數(shù)和最大的那個(gè)正數(shù)相乘
- 如果只有三個(gè)數(shù)，直接返回三個(gè)數(shù)的乘積
- 將nums有小到大排序，如果nums中全為正數(shù)，則取最大三個(gè)數(shù)相乘
- 如果nums中有負(fù)數(shù)，則判斷最小的兩個(gè)數(shù)乘積nums[0] * nums[1]和倒數(shù)第二，第三大的兩個(gè)數(shù)nums[len1 - 2] * nums[len1 - 3]乘積的大小，選較大的和最大的數(shù)相乘
代碼
MaximumProduct.go
package _628_Maximum_Product_Three_Numbers
import (
"sort"
"fmt"
)
func MaximumProduct(nums []int) int {
len1 := len(nums)
if len1 == 3 {
return nums[0] * nums[1] * nums[2]
}
sort.Ints(nums)
fmt.Printf("int:%+v\n", nums)
if (nums[0] * nums[1]) > (nums[len1 - 2] * nums[len1 - 3]) {
return nums[0] * nums[1] * nums[len1 - 1]
} else {
return nums[len1 - 1] * nums[len1 - 2] * nums[len1 - 3]
}
}
測試
MaximumProduct_test.go
package _628_Maximum_Product_Three_Numbers
import "testing"
func TestMaximumProduct(t *testing.T) {
var tests = []struct{
input []int
output int
}{
{[]int{1,2,3}, 6},
{
[]int{-710,-107,-851,657,-14,-859,278,-182,-749,718,-640,127,-930,-462,694,969,143,309,904,-651,160,451,-159,-316,844,-60,611,-169,-73,721,-902,338,-20,-890,-819,-644,107,404,150,-219,459,-324,-385,-118,-307,993,202,-147,62,-94,-976,-329,689,870,532,-686,371,-850,-186,87,878,989,-822,-350,-948,-412,161,-88,-509,836,-207,-60,771,516,-287,-366,-512,509,904,-459,683,-563,-766,-837,-333,93,893,303,908,532,-206,990,280,826,-13,115,-732,525,-939,-787},
972256230,
},
}
for _, v := range tests {
ret := MaximumProduct(v.input)
if ret == v.output {
t.Logf("pass")
} else {
t.Errorf("fail, want %+v, get %+v", v.output, ret)
}
}
}
