LeetCode 215. Kth Largest Element in an Array--Divide and Conquer(分治法)

题目链接

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.

解：依然可以用分治法解决，类似于快速排序的思想，在vector中选取一个key值，其余的每一项和key值比较，根据比较结果将其余数分别存入两个vector，然后可以判断所求值是不是key值或在哪个vector中，进而递归求解

重点在于key值的选择，根据数组的规律合理选择key，可以快速找到所求值，减少空间的开销

首先我根据常规将nums[0]选为key值，结果就遇到了Memory Limit Exceeded的问题，将nums[size-1]作为key依然不行，说明测试样例中数的排列规律近似为递增或递减，我试着将nums[size/2]作为key，通过了：

class Solution {public:    int findKthLargest(vector<int>& nums, int k) {        int size = nums.size();        vector<int> temp1;        vector<int> temp2;        int mid = size/2;        int key = nums[mid];        for (int i = 0; i < size; i++) {            if(i == mid) continue;            if (nums[i] > key) {                temp1.push_back(nums[i]);            }            else {                temp2.push_back(nums[i]);            }        }        int size1 = temp1.size();        if (size1 == k-1) return nums[mid];        else if (size1 < k-1) {            return findKthLargest(temp2, k-1-size1);        }        else {            return findKthLargest(temp1, k);        }    }};