Leetcode 0215: 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.

Example 1:

Input: [3,2,1,5,6,4] and k = 2
Output: 5

Example 2:

Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4

Example 3:

Input: s = “A”, numRows = 1
Output: “A”

Note:

You may assume k is always valid, 1 ≤ k ≤ array’s length.

Time complexity: O(n)
快速排序的思想：

1. 在给定区间[start, end]中，选中某个数x，将大于等于x的放在左边，小于x的放在右边，其中[start, j]是大于等于x的区间，[j + 1,end]是小于x的区间
2. 判断出第k大与j的大小关系，若符合大于等于x，则递归到[start, j]区间，否则递归到[j + 1,end]的区

class Solution {
    
    
    int[] nums;
    public int findKthLargest(int[] nums, int k) {
    
    
        this.nums = nums;
        return quick(0, nums.length-1, nums.length - k);
    }
    
    private int quick(int start, int end, int k){
    
    
        if(start >= end) return nums[end];
        int mid = start + (end - start)/2;
        int pivot = nums[mid];
        int l = start;
        int r = end;
        while(l <= r){
    
    
            while(l <= r && nums[l] < pivot){
    
    
                l++;
            }
            while(l <= r && nums[r] > pivot){
    
    
                r--;
            }
            if(l <= r){
    
    
                swap(l++, r--);
            }
        }
        
        if(l <= k){
    
    
            return quick(l, end, k);
        }
        if(k <= r){
    
    
            return quick(start, r, k);
        }
        return nums[k];
    }
    
    private void swap(int l, int r){
    
    
        int temp = nums[l];
        nums[l] = nums[r];
        nums[r] = temp;
    }
}