Kth Largest Element in an Array

题目描述：

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.

这个题借助快排的思想，先复习下快速排序吧！代码如下：

private int partition(int[] nums, int p, int r) {  
    print(nums);
    int x = nums[r];  

    int i = p-1;   
    for(int j=p;j<r;j++){  
        if(nums[j]<=x) {  
            i++;
            swap(nums, i, j); 
        }   
    }  
    swap(nums,i+1, r);
    return i+1;  
} 
private void swap(int[] nums, int i, int j) {  
    int temp = nums[i];  
    nums[i] = nums[j];  
    nums[j] = temp;  
}  

public void qSort(int[] a,int s,int t){
    if(s<t){
        int privotkey = partition(a, s, t);
        qSort(a, s, privotkey-1);
        qSort(a, privotkey+1, t);
    }
}

public void quickSort(int[] a){
    qSort(a, 0, a.length-1);
}

这里的关键部分是partition过程，实现对nums[p...r]的原址排序，将小于nums[r]的排在左边，大于nums[r]的排在右边，返回nums[r]在数组中的应排的位置。

这个题借用此partition，代码如下：

public class Solution {
    public int findKthLargest(int[] nums, int k) {  
        return findK(nums, nums.length-k, 0, nums.length-1);  
    }  
      
    private int findK(int[] nums, int k, int i, int j) {  
        if(i>=j) return nums[i];  
        int m = partition(nums, i, j);  
        if(m==k) return nums[m];  
        else if(m<k) {  
            return findK(nums, k, m+1, j);  
        } else {  
            return findK(nums, k, i, m-1);  
        }  
    }  
    
    private void swap(int[] nums, int i, int j) {  
        int temp = nums[i];  
        nums[i] = nums[j];  
        nums[j] = temp;  
    }  
    
    private int partition(int[] nums, int p, int r) {  
        int x = nums[r];  
        int i = p-1;   
        
        for(int j=p;j<r;j++){  
            if(nums[j]<=x) {  
            	i++;
                swap(nums, i, j); 
            }   
        }  
        swap(nums,i+1, r);
        return i+1;  
    } 
}