4. Median of Two Sorted Arrays

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本文介绍了一种在O(log(m+n))的时间复杂度下找到两个有序数组中位数的方法。通过双指针技术将两个数组合并，并根据合并后的数组长度计算中位数。适用于竞赛编程和算法面试。

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4. Median of Two Sorted Arrays

Hard

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)).

You may assume nums1 and nums2 cannot be both empty.

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

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这里使用了双指针进行排序

#include<bits/stdc++.h>
using namespace std;
class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2){
    	int n1=nums1.size();
    	int n2=nums2.size();
    	int i=0,j=0;
    	vector<int> ans;
    	while(i<n1||j<n2){//双指针进行排序
    		if(i<n1&&j<n2){//两个vector都没排完
    			if(nums1[i]<=nums2[j]){
    				ans.push_back(nums1[i]);
    				i++;
    			}else{
    				ans.push_back(nums2[j]);
    				j++;
    			}
    		}else if(i<n1&&j>=n2){//一个vector排完了，另一个没排完
    			ans.push_back(nums1[i]);i++;
    		}else if(i>=n1&&j<n2){
    			ans.push_back(nums2[j]);j++;
    		}
    	}
    	if((n1+n2)%2!=0){
    		return (double)ans[(n1+n2+1)/2-1];
    	}else{
    		return (double)(ans[(n1+n2)/2]+ans[(n1+n2)/2-1])/2;
    	}
    }
};