找第K小个数

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快排进行划分：

import java.util.ArrayList;import java.util.Arrays;import java.util.List;public class Solution {public static void main(String[] args){int[] nums = {2,4,3,6,1,5};Solution s = new Solution();int k = 3;int res = s.findkminNum(k,0,5,nums);System.out.println(res);}    public int findkminNum(int k,int start,int end,int[] nums){    if(k==0||k>end-start+1||end<start) return -1;    int key = nums[end];        int i = start-1;    for(int j=start;j<end;j++){    if(nums[j]<key){    i++;    int tmp = nums[i];    nums[i] = nums[j];    nums[j] = tmp;    }    }    i++;int tmp = nums[i];nums[i] = nums[end];nums[end] = tmp;        int partition = i;    if(partition-start+1==k) return nums[partition];    else if(partition-start+1>k)  return findkminNum(k,start,partition-1,nums);    else  return findkminNum(k-(partition-start+1),partition+1,end,nums);    }}

讲解得比较详细的博客： http://www.yalewoo.com/quickselect_and_linearselect.html

下面这种方法思路"巧妙"？个人觉得很不够直观，编程到一半比较懒，暂且记录下来，挖个坑

import java.util.ArrayList;import java.util.Arrays;import java.util.List;public class Solution {public static void main(String[] args){int[] nums = {2,4,3,6,1,5};Solution s = new Solution();int k = 3;int res = s.findkminNum(k,nums);System.out.println(res);}    public int findkminNum(int k, int[] nums){    int n = nums.length;    int sub = n/5;    for(int i=0;i<sub;i++){    int[] elem = new int[5];    for(int j=0;j<5;j++){    if(5*i+j<n)elem[j]=nums[5*i+j];    /*     * insert sort ---  start     */    int pos = j;    int tmp = nums[pos];    while(j>0&&elem[j-1]>elem[pos]){    j--;    }    for(int h=pos;h>j;h--){    elem[h]=elem[h-1];    }    elem[j] = tmp;    /*     * insert sort ---  end     */    }    for(int j=0;j<5;j++){    if(5*i+j<n) nums[5*i+j] = elem[j];    }    }            int[] mediumArray = new int[sub];    for(int i=0;i<sub-1;i++){    mediumArray[i] = nums[5*i+2];    }    if(n%5>=3||n%5==0) mediumArray[sub-1] = nums[5*(sub-1)+2];    if(n%5<3||n%5!=0)  mediumArray[sub-1] = nums[n-1];        int partition = findkminNum((sub+1)/2,mediumArray);    int countSmaller = 3*((sub+1)/2);    int countBigger  = 3*(n-(sub+1)/2);    int partition2    if(k==partition)        return k;    }}

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