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.

思路一：  利用top-k的思想，这里可以用优先队列或者muiti-set来实现。O（Nlogk）

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        priority_queue<int> pq(nums.begin(), nums.end());
        for (int i = 0; i < k - 1; i++)
            pq.pop(); 
        return pq.top();
    }
};

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        multiset<int> mset;
        int n = nums.size();
        for (int i = 0; i < n; i++) { 
            mset.insert(nums[i]);
            if (mset.size() > k)
                mset.erase(mset.begin());
        }
        return *mset.begin();
    }
};

思路二： 基于partition 函数的思路，算法复杂度O(N)

class Solution { 
public:
    int partition(vector<int>& nums, int left, int right) {
        int pivot = nums[left];
        int l = left + 1, 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 findKthLargest(vector<int>& nums, int k) {
        int left = 0, right = nums.size() - 1;
        while (true) {
            int pos = partition(nums, left, right);
            if (pos == k - 1) return nums[pos];
            if (pos > k - 1) right = pos - 1;
            else left = pos + 1;
        }
    }
};

 follow up :

如果是要求第三大的数？

利用三个指针，基于top-3 的思想。