【LeetCode】628.Maximum Product of Three Numbers(Easy)解题报告

【LeetCode】628.Maximum Product of Three Numbers(Easy)解题报告

题目地址：https://leetcode.com/problems/maximum-product-of-three-numbers/description/

题目描述：

  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：

  1.The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].

  2.Multiplication of any three numbers in the input won’t exceed the range of 32-bit signed integer.

  题意就是找到三个数最大的乘积，但要考虑可能有负数。两种解法。

Solution1:

//三个数成绩最大
//绝对值最大：1.三正数 2.一正数，两负数
//time:O(nlogn)
//space:O(nlogn)
class Solution {
public int maximumProduct(int[] nums) {
Arrays.sort(nums);
return Math.max(nums[nums.length-1]*nums[nums.length-2]*nums[nums.length-3],nums[0]*nums[1]*nums[nums.length-1]);
}
}

Solution2：

//三个数成绩最大
//绝对值最大：1.三正数 2.一正数，两负数
//time:O(n)
//space:O(1)
class Solution {
public int maximumProduct(int[] nums) {
int min1 = Integer.MAX_VALUE;
int min2 = Integer.MAX_VALUE;
int max1 = Integer.MIN_VALUE;
int max2 = Integer.MIN_VALUE;
int max3 = Integer.MIN_VALUE;
for(int num : nums){
if(num <= min1){ //min1 最小
min2 = min1;
min1 = num;
}else if(num <= min2){ //min2 第二小
min2 = num;
}
if(num >= max1){  //max1 最大
max3 = max2;
max2 = max1;
max1 = num;
}else if(num >= max2){ //max2 第二大
max3 = max2;
max2 = num;
}else if(num >= max3){ //max3 第三大
max3 = num;
}
}
return Math.max(min1*min2*max1,max1*max2*max3);
}
}

Date：2018年2月11日