[leetcode]Kth Smallest Element in an Array

利用QuickSelect的思想进行递归查找

QuickSelect:https://en.wikipedia.org/wiki/Quickselect

主要思想是借鉴QuickSort的分割思路，不同的是，QuickSelect只从一边进行遍历，以最后一个元素为pivot，将原无序数组分割为两部分

The basic idea is to use Quick Select algorithm to partition the array with pivot:

Put numbers < pivot to pivot's leftPut numbers > pivot to pivot's right

Then

if indexOfPivot == k, return A[k]else if indexOfPivot < k, keep checking left part to pivotelse if indexOfPivot > k, keep checking right part to pivot

Time complexity = O(n)

利用QuickSelect查找第k大的数，同理可以查找第k小的数

public class Solution {    public int findKthLargest(int[] nums, int k) {       return findKthLargest(nums,0,nums.length-1,k-1);     }    public int findKthLargest(int[] nums,int start,int end,int k){        if(start>end){            return Integer.MAX_VALUE;        }        int left = start;        int pivotValue = nums[end];        for(int i=start;i<end;i++){            if(nums[i]>pivotValue){                if(left!=i){                    swap(nums,left,i);                }                left++;            }        }        swap(nums,left,end);        if(left==k){            return nums[left];        }else if(left<k){            return findKthLargest(nums,left+1,end,k);        }else            return findKthLargest(nums,start,left-1,k);    }    public void swap(int[] nums,int i,int j){        int temp=nums[i];        nums[i] = nums[j];        nums[j] = temp;    }}