628. Maximum Product of Three Numbers

题目

628. Maximum Product of Three Numbers

解题思路

看了discuss才AC的……一开始想到直接用快排排序选出前3个最大的数字，但遇到了有负数的情况就WA，然后去看了discuss才知道也要选出最小的两个数字（可能都是负数可以负负得正），最后再比较哪3个乘起来大就选哪个。

我的思路缺陷在于没有考虑到有负数的情况和负负得正的定理。

具体代码

快排大法好……两年死活记不住的快排代码这几天写到记住了真是熟能生巧=-=

class Solution {public:    void swap(vector<int>& nums, int left, int right) {        int t = nums[left];        nums[left] = nums[right];        nums[right] = t;    }    int partition(vector<int>& nums, int left, int right) {        int privot = nums[right];        int low = left, high = right - 1;        while (low < high) {            while (low < high && nums[low] <= privot)                ++low;            while (low < high && nums[high] > privot)                --high;            if (low < high)                swap(nums, low, high);        }        if (nums[high] <= privot)            swap(nums, ++high, right);        else            swap(nums, high, right);        return high;    }    void quicksort(vector<int>& nums, int left, int right) {        if (left < right) {            int index = partition(nums, left, right);            quicksort(nums, left, index - 1);            quicksort(nums, index + 1, right);        }    }    int maximumProduct(vector<int>& nums) {        quicksort(nums, 0, nums.size() - 1);        int a = nums[nums.size() - 1] * nums[nums.size() - 2] * nums[nums.size() - 3];        int b = nums[nums.size() - 1] * nums[0] * nums[1];        return a > b ? a : b;    }};