[leetcode]215. Kth Largest Element in an Array

divide and conquer

题目：

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.

解法：

分而治之：

利用快速排序，pivot是子数组的第一个元素。大于关键点pivot的值放在左边，小于关键点pivot的值放在右边。如果关键点恰好是nums[k-1]，那么该关键点就是数组的第k大的数。如果关键点

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

优先队列：

优先队列可以从大到小自动排列，所以将前k-1个数据弹出，队头就是第k大数据。

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