求两个排好序的数组中的第k小数字

  最近各种原因一直看论文做ppt，觉得还是得养成刷题的习惯，保持思维的活跃度，而且可以多手准备。

  很多人推荐LeetCode，第一次玩，OJ的题目太少。去看了讨论版，随便选了道容易上手的，给定两个排好序的数组，找出第k小的数字。

  O( (m+n)log(m+n))的算法应该没人会用吧。

  O(k)的算法应该很多人都能想到。

  O(log(k))的算法有点难想到。不过想通了发现思维也很简单。第k小的数字为x，那么数组1一定有i个数字小于x，数组2一定有j个数字小于x，注意i+j=k-1。因为k已知，所以我们只要搜索i即可。然后我们可以想象到，如果元素array1[i+1]为两个数组的并的第k小的数字的话，那么它满足 arrary1[i+1] >= array2[j] && array1[i+1]<=array2[j+1]。同理如果array2[j+1]为两个数组的并的第k小的数字的话，那么它满足 arrary2[j+1] >= array1[i] && array2[j+1]<=array1[i+1]。二分搜索数组1的i，并对照数组2对应的j，如果array1[i]的值太大，搜索数组1的i的前半区间，否则搜索后半区间。这个二分有点难描述，具体可以参考：

http://leetcode.com/2011/01/find-k-th-smallest-element-in-union-of.html

 

实现了O(k)和O(log(k))的代码加测试代码：

import java.util.*;public class main {//Generate two random arrays.public static void GenerateRanSortedArrs(ArrayList<Integer> arr1, ArrayList<Integer> arr2){Random ranGen = new Random();int n = ranGen.nextInt(100);int m = ranGen.nextInt(100);arr1.add(ranGen.nextInt(10));arr2.add(ranGen.nextInt(10));for(int i=1;i<n;i++)arr1.add(arr1.get(i-1) + ranGen.nextInt(100));for(int i=1;i<m;i++)arr2.add(arr2.get(i-1) + ranGen.nextInt(100));}//For testing, generate two fixed number arrays.public static void GenerateFixedSortedArrs(ArrayList<Integer> arr1, ArrayList<Integer> arr2){int[] a1= new int[]{5,5,5,5,5,5};int[] a2= new int[]{5,5,5,5,5,5,5};for(int i=0;i<a1.length;i++ )arr1.add(a1[i]);for(int i=0;i<a2.length;i++ )arr2.add(a2[i]);}//For testing, show the arraypublic static void ShowArr(ArrayList<Integer> arr){for(int i=0;i<arr.size();i++){System.out.print(arr.get(i) + ", ");}System.out.println();}//O(k) algorithmpublic static int KthValueFromSortedArrs_Linear(ArrayList<Integer> arr1, ArrayList<Integer> arr2, int k){if(k > arr1.size() + arr2.size())return -1;int MAX = 1<<31 -1 ;int MIN = 1<<31;int c =0;for(int i=0;i<k;)for(int j=0;j<k;){int arr1_i = (i<arr1.size()) ? arr1.get(i) : MAX;int arr2_j = (j<arr2.size()) ? arr2.get(j) : MAX;if(arr1_i < arr2_j){c++;if(c==k)return arr1_i;i++;}else if(arr2_j < arr1_i){c++;if(c==k)return arr2_j;j++;}else if(arr1_i == arr2_j){c+=2;if(c>=k)return arr1_i;i++;j++;}}return -1;}//O(log(k)) algorithmpublic static int KthValueFromSortedArrs_Log(ArrayList<Integer> arr1, ArrayList<Integer> arr2, int k){if(k > arr1.size() + arr2.size())return -1;int MAX = 1<<31-1;int MIN = 1<<31;int i,j;//i+j == k-1int left=0,right=k-1;while(true){if(right >= left)i = (left+right)/2;elsei = -1;j = k-1-1-1-i;if(i >= arr1.size())//it means there is not enough i{right = i - 1;continue;}int arr1_i = (i>=0)? arr1.get(i) : MIN; int arr1_i_1 = (i+1<arr1.size())? arr1.get(i+1) : MAX;if(j >= arr2.size())//it means there is no enough j, so i must get bigger.{left = i + 1;continue;}int arr2_j = (j>=0)? arr2.get(j) : MIN;int arr2_j_1 = (j+1<arr2.size())? arr2.get(j+1) : MAX;if (arr1_i_1 >= arr2_j&& arr1_i_1 <= arr2_j_1)return arr1_i_1;else if (arr2_j_1 >= arr1_i&& arr2_j_1 <= arr1_i_1)return arr2_j_1;else if (arr1_i > arr2_j) {right = i - 1;}else if (arr1_i < arr2_j) {left = i+1;}}}public static void main(String[] args){for(int i=0;i<1000;i++){ArrayList<Integer> arr1 = new ArrayList<Integer>();ArrayList<Integer> arr2 = new ArrayList<Integer>();GenerateRanSortedArrs(arr1,arr2);for(int k=3;k<=arr1.size()+arr2.size();k++){      int k1 = KthValueFromSortedArrs_Linear(arr1,arr2,k) ;      int k2 = KthValueFromSortedArrs_Log(arr1,arr2,k) ;      //System.out.println(k1 + " " + k2);      if(k1 != k2)      {     ShowArr(arr1);     ShowArr(arr2);     System.out.println(k1 + " " + k2);      }}}System.out.println("Finished Running All The Demos!"); }}