image.png

1. 解法一，排序然后返回第k大
   重点是降序的表示方法： greater<int> ()

class Solution {
public:
    /*
     * @param n: An integer
     * @param nums: An array
     * @return: the Kth largest element
     */
    bool cmp(int a, int b){
            return a > b;
        }
    int kthLargestElement(int n, vector<int> &nums){
        // write your code here
        sort(nums.begin(), nums.end(), greater<int> ());
        return nums[n-1];
    }
};

或者:

2. heapsort
   这个解法在lintcode上一直报错，leetcodeAC
3. Kth Largest Element in an Array

class Solution {
public:
    void heapify(vector<int> &nums, int heapsize, int index){
        int largest = index;
        int left = 2 * largest + 1;
        int right = 2 * largest + 2;
        if(nums[left] > nums[largest] && left < heapsize){
            largest = left;
        }
        if(nums[right] > nums[largest] && right < heapsize){
            largest = right;
        }
        if(largest != index){
            swap(nums[index], nums[largest]);
            heapify(nums, heapsize, largest);
        }
        
    }
    
    int findKthLargest(vector<int>& nums, int k) {
        int hp_size = nums.size();
        for(int i = hp_size/2 - 1; i >= 0; i--){
            heapify(nums, hp_size, i);
        }
        for(int i = 0; i < k; i++){
            swap(nums[0], nums[hp_size - 1]);
            hp_size--;
            heapify(nums, hp_size, 0);
        }
        return nums[hp_size];
    }
    
};

3. quickselect

class Solution {
public:
    /*
     * @param n: An integer
     * @param nums: An array
     * @return: the Kth largest element
     */
    int partition(vector<int> &nums, int left, int right){
        int pivot = nums[left];
        int l = left + 1;
        int r = right;
        while(l <= r){
            if(nums[l] < pivot && nums[r] > pivot){
                swap(nums[l++],nums[r--]);
            }
            if(nums[l] >= pivot ){
                l++;
            }
            if(nums[r] <= pivot  ){
                r--;
            }
        }
        swap(nums[left], nums[r]);
        return r;
    }
    int kthLargestElement(int n, vector<int> &nums) {
        // write your code here
        int left = 0;
        int right = nums.size() - 1;
        while(1){
            int index = partition(nums,left,right);
            if(index == n-1){
                return nums[n-1];
            }
            if(index > n - 1){
                right = index - 1;
            }
            else{
                left = index + 1;
            }
        }
    }
};

4. priority_queue

class Solution {
public:
    /*
     * @param n: An integer
     * @param nums: An array
     * @return: the Kth largest element
     */
    int kthLargestElement(int n, vector<int> &nums) {
        // write your code here
        priority_queue<int> pq(nums.begin(),nums.end());
        for(int i = 0 ; i < n - 1; i++){
            pq.pop();
        }
        return pq.top();
    }
};