三種方法找到數組中的第k個最大元素【此後無良辰】

沒想到今天考試竟然遇到了原題第k小，其實就是n-k+1大元素（本篇之前寫的放私密了）

分界線------------------------------------------------------------------------------------分界線

題面描述

基本思路

1.排序返回值

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end());
        return nums[nums.size()-k];
    }
};

2.根據快速排序進行分治

class Solution {
public:
    int quickSelect(vector<int>& a, int l, int r, int index) {
        int q = randomPartition(a, l, r);
        if (q == index) {
            return a[q];
        } else {
            return q < index ? quickSelect(a, q + 1, r, index) : quickSelect(a, l, q - 1, index);
        }
    }

    inline int randomPartition(vector<int>& a, int l, int r) {
        int i = rand() % (r - l + 1) + l;
        swap(a[i], a[r]);
        return partition(a, l, r);
    }

    inline int partition(vector<int>& a, int l, int r) {
        int x = a[r], i = l - 1;
        for (int j = l; j < r; ++j) {
            if (a[j] <= x) {
                swap(a[++i], a[j]);
            }
        }
        swap(a[i + 1], a[r]);
        return i + 1;
    }

    int findKthLargest(vector<int>& nums, int k) {
        srand(time(0));
        return quickSelect(nums, 0, nums.size() - 1, nums.size() - k);
    }
};

3.根據堆排序的性質

class Solution {
public:
    void maxHeapify(vector<int>& a, int i, int heapSize) {
        int l = i * 2 + 1, r = i * 2 + 2, largest = i;
        if (l < heapSize && a[l] > a[largest]) {
            largest = l;
        } 
        if (r < heapSize && a[r] > a[largest]) {
            largest = r;
        }
        if (largest != i) {
            swap(a[i], a[largest]);
            maxHeapify(a, largest, heapSize);
        }
    }

    void buildMaxHeap(vector<int>& a, int heapSize) {
        for (int i = heapSize / 2; i >= 0; --i) {
            maxHeapify(a, i, heapSize);
        } 
    }

    int findKthLargest(vector<int>& nums, int k) {
        int heapSize = nums.size();
        buildMaxHeap(nums, heapSize);
        for (int i = nums.size() - 1; i >= nums.size() - k + 1; --i) {
            swap(nums[0], nums[i]);
            --heapSize;
            maxHeapify(nums, 0, heapSize);
        }
        return nums[0];
    }
};

提交oj

從下到上爲1–》3的方法一一對應