1.求两个已经排好序的数组的中位数
2.这道题目与leetcode中的Median of Two Sorted Arrays相似，只是偶数情况取前一个，不用求平均
3.采用二分法进行查找。
数组分为如下部分：
{a[0],a[1],a[2],….a[i-1] | a[i],a[i+1],…a[m-1]}
{b[0],b[1],b[2],….b[j-1] | b[j],b[j+1],…b[n-1]}
然后根据a[i-1]与b[j]、a[i]与b[j-1]的大小，进行二分查找。
4.注意边界条件，可以看代码的注释。
Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.
Given two increasing sequences of integers, you are asked to find their median.
Input
Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (<=1000000) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.
Output
For each test case you should output the median of the two given sequences in a line.
Sample Input
4 11 12 13 14 5 9 10 15 16 17
Sample Output
13
AC代码：
[c language=”++”]
//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
//#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;
/*
0
0
1 100
0
0
1 100
1 10
1 100
1 10
2 20 30
2 10 20
2 30 40
3 10 20 30
3 40 50 60
4 10 20 30 70
4 40 50 60 80
3 10 20 30
4 40 50 60 80
2 3 4
0
*/
/*
{a[0],a[1],a[2],….a[i-1] | a[i],a[i+1],…a[m-1]}
{b[0],b[1],b[2],….b[j-1] | b[j],b[j+1],…b[n-1]}
*/
int findMedianSortedArrays(int *nums1, int*nums2,int m,int n) {
if (m>n) return findMedianSortedArrays(nums2, nums1,n,m);
int minIdx = 0, maxIdx = m;//第一个数组的范围
int i, j;//两个数组的下标
int num1, num2;
int mid = (m + n + 1) >> 1;//为什么+1
while (minIdx <= maxIdx)
{
i = (minIdx + maxIdx) >> 1;//取中间值
j = mid – i;
if (i<m && j>0 && nums2[j – 1] > nums1[i])
minIdx = i + 1;
else if (i>0 && j<n && nums2[j] < nums1[i – 1])
maxIdx = i – 1;
else
{
if (i == 0) num1 = nums2[j – 1];
else if (j == 0) num1 = nums1[i – 1];
else num1 = max(nums1[i – 1], nums2[j – 1]);
break;
}
}
return num1;
}
int main(void)
{
int m, n;
scanf("%d", &m);
int *nums1 = new int[m];
for (int i = 0; i < m; i++)
{
scanf("%d", &nums1[i]);
}
scanf("%d", &n);
int *nums2 = new int[n];
for (int i = 0; i < n; i++)
{
scanf("%d", &nums2[i]);
}
//cout << findMedianSortedArrays(nums1, nums2, m, n) << endl;
int *a, *b;
if (m > n)
{
a = nums2;
b = nums1;
swap(m, n);
}
else
{
a = nums1;
b = nums2;
}
/*
{a[0],a[1],a[2],….a[i-1] | a[i],a[i+1],…a[m-1]}
{b[0],b[1],b[2],….b[j-1] | b[j],b[j+1],…b[n-1]}
*/
int ans = INT_MIN;
bool getAns = false;
int l = 0, r = m;//必须是r=m，数组a是长度较小的一个，假如长度为0，那么m-1=-1，下面会直接出错
int i = (l + r) / 2;
int j = (m + n + 1) / 2 – i;
while (l <= r)//同理，假如m==0，至少确保循环执行了一次
{
i = (l + r) / 2;//假如m为0，i为0
j = (m + n + 1) / 2 – i;//+1很重要
if (i > 0 && j<n && a[i – 1]>b[j])
r = i – 1;
else if (j > 0 && i<m && b[j – 1]>a[i])
l = i + 1;
else
{
if (i == 0)
ans = b[j – 1];
else if (j == 0)
ans = a[i – 1];
else
ans = max(a[i – 1], b[j – 1]);
break;
}
}
cout << ans << endl;
return 0;
}
[/c]