004 Median of Two Sorted Arrays 两个有序数组的中位数

There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0

Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5

详见：https://leetcode.com/problems/median-of-two-sorted-arrays/description/

Java实现：

方法一：常规解法

import java.util.Arrays;
class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int size1=nums1.length;
        int size2=nums2.length;
        int size=size1+size2;
        int[] nums = new int[size];
        for(int i = 0; i<size1; i++){
            nums[i] = nums1[i];
        }
        for(int i = size1; i<size; i++){
            nums[i] = nums2[i-size1];
        }
        Arrays.sort(nums);
        double median;
        if((size-1)%2 == 0){
            median = nums[(size-1)/2] * 1.0;
        } else {
            median = (nums[(size-1)/2] + nums[((size-1)/2)+1])/2.0;
        }
        return median;
    }
}

　方法二：求第k小的数

public class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int size1 = nums1.length;
        int size2 = nums2.length;
        int size = size1 + size2;
        if(size % 2 == 1){
            return findKth(nums1, 0, size1, nums2, 0, size2, size / 2 + 1);
        }else{
            return (findKth(nums1, 0, size1, nums2, 0, size2, size / 2) + findKth(nums1, 0, size1, nums2, 0, size2, size / 2 + 1)) /2;
        }
    }
    public double findKth(int[] nums1, int start1, int size1, int[] nums2, int start2, int size2, int k){
        if(size1 - start1 > size2 -start2){
            return findKth(nums2, start2, size2, nums1, start1, size1, k);
        }
        if(size1 - start1 == 0){
            return nums2[k - 1];
        }
        if(k == 1){
            return Math.min(nums1[start1], nums2[start2]); // k==1表示已经找到第k-1小的数，下一个数为两个数组start开始的最小值
        }
        int p1 = start1 + Math.min(size1 - start1, k / 2); // p1和p2记录当前需要比较的那个位
        int p2 = start2 + k - p1 + start1;
        if(nums1[p1 - 1] < nums2[p2 - 1]){
            return findKth(nums1,  p1, size1, nums2, start2, size2, k - p1 + start1);
        }else if(nums1[p1 - 1] > nums2[p2 -1]){
            return findKth(nums1, start1, size1, nums2, p2, size2, k - p2 + start2);
        }else{
            return nums1[p1 - 1];
        }
    }
}

C++实现：

 class Solution {
 public:
     double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
         int size1=nums1.size();
         int size2=nums2.size();
         int size=size1+size2;
         if(size&0x1)
             return findKTh(nums1.begin(),size1,nums2.begin(),size2,size/+);
         else
             return (findKTh(nums1.begin(),size1,nums2.begin(),size2,size/)+findKTh(nums1.begin(),size1,nums2.begin(),size2,size/+))/;
     }
     double findKTh(vector<int>::iterator nums1,int m,vector<int>::iterator nums2,int n,int k)
     {
         if(m>n)
             return findKTh(nums2,n,nums1,m,k);
         if(m==)
             return *(nums2+k-);
         if(k==)
             return min(*nums1,*nums2);
         int pa=min(k/,m),pb=k-pa;
         if(*(nums1+pa-)<*(nums2+pb-))
             return findKTh(nums1+pa,m-pa,nums2,n,k-pa);
         else if(*(nums1+pa-)>*(nums2+pb-))
             return findKTh(nums1,m,nums2+pb,n-pb,k-pb);
         else
             return *(nums1+pa-);
     }
 };

参考：

https://www.cnblogs.com/leavescy/p/5877627.html

https://www.cnblogs.com/bakari/p/5082155.html

http://blog.csdn.net/yutianzuijin/article/details/11499917/