【leetcode】215. Kth Largest Element in an Array

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.

解题思路：最简单的，我们可以选择将数组排序，然后返回第n-k项。另一种方法，我将数组以中间的数大小m为基准将数组分为左右两个数组，左数组小于m,右数组大于m；若右数组的大小q仍然大于k,则继续对右数组进行划分；当右数组大小等于k时，则该数组最小值即为所求，若右数组大小等于k-1,则划分数m即为所求；若右数组小于k，则将k = k - q -1；对左数组进行划分。

解答：

class Solution {

public:

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

    int num;

        find(nums,k,num);

        return num; 

    }

private:

void find(vector<int>& nums,int k,int& num){
vector<int> a,b;
int n = nums.size();
int r = n / 2;
int m = nums[r];
nums[r] = nums[0];
nums[0] = m;
for(int i = 1;i < n; i ++){

if(nums[i] > m){
b.push_back(nums[i]);
}
else{
a.push_back(nums[i]);
}
}
int q = b.size();
int w = a.size();
if(q > k){
find(b,k,num);
}
else if(q == k){
int e = b[0];
for(int i= 0;i < q;i ++){

    if(b[i] < e){

    e = b[i];
  }
  }
  num = e;
}
else if(q == k -1){
num = m;
}
else{
k = k - q -1;
if(w == k){
int e = a[0];
for(int i= 0;i < w;i ++){

    if(a[i] < e){

    e = a[i];
  }
  }
  num = e;
}
else{
find(a,k,num);
}
}

}

};