[leetcode-628-Maximum Product of Three Numbers]

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.

思路：

首先排序，然后分别判断数组元素最大值是正是负情况。

  int maximumProduct(vector<int>& nums)
  {

    sort(nums.begin(),nums.end());
    int len = nums.size();

    int a,b,c;
    c = nums[len-];
    b = nums[len-];
    a = nums[len-];
    if(a>)return  max(nums[]*nums[]*c,a*b*c);
    else if( a == )
    {
      if(len==)return ;
      if(len>=)return nums[len-]*nums[len-]*c;//l两个负数
      else return a*b*c;
    }
    else if(a<)
    {
      if(c< )return a*b*c;
      if(c>= &&b< )return nums[]*nums[]*c;
      if(c>= && b> &&len>=)return nums[]*nums[]*c;
      if(c>= && b> &&len==)return a*b*c;
    }

    return ;
  }
  

感觉写出来 超级啰嗦 惨不忍睹，于是看到了如下代码，

醍醐灌顶，五体投地。 排序过后，依次讨论前三个，后三个，以及后两个跟第一个，前两个跟最后一个。

 public int maximumProduct(int[] nums) {
            Arrays.sort(nums);
            int n = nums.length;
            int s = nums[n-] * nums[n-] * nums[n-];
            s = Math.max(s, nums[n-] * nums[n-] * nums[]);
            s = Math.max(s, nums[n-] * nums[] * nums[]);
            s = Math.max(s, nums[] * nums[] * nums[]);
            return s;
        }