4. Median of Two Sorted Arrays（求两个有序数组的中位数）

题目描述：

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

知识补充

1.vector类的使用

vector可以看做一个类，当我们不确定数组的大小时，vector可以看做动态的数组。

vector初始化

vector <int> a；//大小没有定义，可以动态的添加删除 vector <int> b(10);//初始化含有10个初始值为0的vector vector <int> c(10,1);//初始化含有10个初始值为1的vector vector <string> d;//也可以初始化字符串型数组 int arr1[]={0,5,7,8,13}； vector< int > nums1(arr1,arr1+5);

输出vector中的元素

vector<int>::iterator it;//it为迭代器 for( it=arr1.begin();it!=arr1.end();++it) { cout<<*it<<" ";//注意it前有个*号 }

将两个vector合并为一个

vector< int > nums3; nums3.insert(nums1.end(),nums2.begin(),nums2.end());

测试代码

class Solution { public: double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { vector< int > nums3; nums3.insert(nums3.end(),nums1.begin(),nums1.end()); nums3.insert(nums3.end(),nums2.begin(),nums2.end());//把两个数组组合成新数组 std::sort (nums3.begin(), nums3.end());//对数组进行排序 float media = 0.0; int pos = nums3.size(); if(pos%2==0){ media = static_cast<float>((nums3[pos/2]+nums3[(pos/2)-1]))/2;//如果长度为偶数取两数平均值 }else { media = nums3[pos/2];//奇数则取中间的数 } return media; } };

性能

参考答案

class Solution { public: double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { int N1 = nums1.size(); int N2 = nums2.size(); if (N1 < N2) return findMedianSortedArrays(nums2, nums1); // Make sure A2 is the shorter one. if (N2 == 0) return ((double)nums1[(N1-1)/2] + (double)nums1[N1/2])/2; // If A2 is empty int lo = 0, hi = N2 * 2; while (lo <= hi) { int mid2 = (lo + hi) / 2; // Try Cut 2 int mid1 = N1 + N2 - mid2; // Calculate Cut 1 accordingly double L1 = (mid1 == 0) ? INT_MIN : nums1[(mid1-1)/2]; // Get L1, R1, L2, R2 respectively double L2 = (mid2 == 0) ? INT_MIN : nums2[(mid2-1)/2]; double R1 = (mid1 == N1 * 2) ? INT_MAX : nums1[(mid1)/2]; double R2 = (mid2 == N2 * 2) ? INT_MAX : nums2[(mid2)/2]; if (L1 > R2) lo = mid2 + 1; // A1's lower half is too big; need to move C1 left (C2 right) else if (L2 > R1) hi = mid2 - 1; // A2's lower half too big; need to move C2 left. else return (max(L1,L2) + min(R1, R2)) / 2; // Otherwise, that's the right cut. } return -1; } };

性能