[LeetCode]628. Maximum Product of Three Numbers

[LeetCode]628. Maximum Product of Three Numbers

题目描述

思路

简单的做法，扫描一次即可，维护两个最小值和三个最大值

看到这个题目，想起一道面试题，一个无法载入内存的数组，从中选n个最大的数
解法，建立一个小根堆，限定元素个数为n，每次插入一个新元素后弹出堆顶元素，扫描一次，堆中的n个元素即为n个最大的数。

这道题思路可以借鉴这种方式，顺便回顾一下调整堆的方法
维护n = 3的小根堆 和 n = 2 的大根堆
最后将小根堆使用大根堆调整，这样最大的元素就是堆顶元素

详见代码

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    void print(vector<int> nums) {
        for (int num : nums)
            cout << num << ' ';
        cout << endl;
    }

    int maximumProduct(vector<int>& nums) {
        vector<int> maxHeap, minHeap;
        for (int i = 0; i < nums.size(); ++i) {
            if (i < 3)
                minHeap.push_back(nums[i]);
            else {
                minHeap.push_back(nums[i]);
                adjustMinHeap(minHeap, 0);
                swap(minHeap[0], minHeap[3]);
                minHeap.pop_back();
            }
            if (i < 2)
                maxHeap.push_back(nums[i]);
            else {
                maxHeap.push_back(nums[i]);
                adjustMaxHeap(maxHeap, 0);
                swap(maxHeap[0], maxHeap[2]);
                maxHeap.pop_back();
            }
        }
        adjustMaxHeap(minHeap, 0);
        print(minHeap);
        print(maxHeap);
        return max(minHeap[0] * minHeap[1] * minHeap[2], maxHeap[0] * maxHeap[1] * minHeap[0]);
    }

    void adjustMaxHeap(vector<int>& nums, int start) {
        int leftChild = 2 * start + 1, rightChild = 2 * start + 2;
        if (leftChild < nums.size()) {
            adjustMaxHeap(nums, leftChild);
            if (nums[start] < nums[leftChild]) {
                swap(nums[start], nums[leftChild]);
                adjustMaxHeap(nums, leftChild);
            }
        }
        if (rightChild < nums.size()) {
            adjustMaxHeap(nums, rightChild);
            if (nums[start] < nums[rightChild]) {
                swap(nums[start], nums[rightChild]);
                adjustMaxHeap(nums, rightChild);
            }
        }
    }

    void adjustMinHeap(vector<int>& nums, int start) {
        int leftChild = 2 * start + 1, rightChild = 2 * start + 2;
        if (leftChild < nums.size()) {
            adjustMinHeap(nums, leftChild);
            if (nums[start] > nums[leftChild]) {
                swap(nums[start], nums[leftChild]);
                adjustMinHeap(nums, leftChild);
            }
        }
        if (rightChild < nums.size()) {
            adjustMinHeap(nums, rightChild);
            if (nums[start] > nums[rightChild]) {
                swap(nums[start], nums[rightChild]);
                adjustMinHeap(nums, rightChild);
            }
        }
    }
};

int main() {
    vector<int> nums = {9,1,5,6,7,2};
    Solution s;

    cout << s.maximumProduct(nums) << endl;

    system("pause");
    return 0;
}