Leetcode[215]-Kth Largest Element in an Array

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given[3,2,1,5,6,4]and k = 2, return 5.

Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

法一：使用STL的sort排序O(NlogN)

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

法二：自己写快速排序O（NlogN）

int findKthLargest(vector<int>& nums, int k){

    quickSort(nums, 0 ,nums.size());
    return nums[nums.size()-k];

}
void quickSort(vector<int> &nums, int left, int right){
    int i = left, j = right - 1;
    if(i < j){
        int po = nums[i];
        while(i < j){
            while(i < j && nums[j] >= po) j--;
            if(i < j) {
                nums[i++] = nums[j];
            }
            while(i < j && nums[i] <= po ) i++;
            if(i < j){
                nums[j--] = nums[i];
            }
        }
        nums[i] = po;
        quickSort(nums, left, i);
        quickSort(nums, i+1, right);

    }
}

法三：使用建堆法 时间复杂度O(klogN)

int findKthLargest(vector<int>& nums, int k){
    make_heap(nums.begin(), nums.end());
    for(auto i=0; i<k-1;i++){
        pop_heap(nums.begin(), nums.end());
        nums.pop_back();
    }
    return nums.front();
}

方法四：此方法是论坛看到的，O（N）

int findKthLargest(vector<int>& nums, int k){
     int i, m, n, pivot, head =0, tail = nums.size()-1, maxV;

    while(1){
        m = head, n= tail;
        pivot = nums[m++];
        while(m <= n) {
            if(nums[m] >= pivot) m++;
            else if(nums[n] < pivot) n--;
            else {
                swap(nums[m++], nums[n--]);
            }
        }
        if(m-head == k) 
            return pivot;
        else if(m-head < k) {
            k -= (m-head); 
            head = m;  
        }
        else {
            tail = m-1;
            head = head+1;
        }
    }

}