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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
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找出两个数组的中位数，
中位数的定义是明确的：按序排好的数组，array.size=n是奇数的话，第（n+1）/2个数字;否则就是第n/2和第（n+1）/2的平均数。
==========思路：
划分思路，
第一步：按照归并思路找到两个数组中的第k大
第二步：按照计算中位数的方法，返回中位数。
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code实现，
class Solution { public: double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { int lengtha = nums1.size(); int lengthb = nums2.size(); int total = lengtha+lengthb; if(total & 0x1){ return find_kth(nums1,nums2,total/2+1); }else{ return (find_kth(nums1,nums2,total/2)+ find_kth(nums1,nums2,total/2+1))/2.0; } } private: int find_kth(vector<int> &A,vector<int> &B,int k){ std::vector<int>::const_iterator p1 = A.begin(); std::vector<int>::const_iterator p2 = B.begin(); int m = 0; while(p1!=A.end() && p2!=B.end()){ if(*p1<=*p2 && m==(k-1)){ return *p1; }else if(*p1>*p2 && m==(k-1)){ return *p2; } if(*p1<=*p2) p1++; else p2++; m++; }//while while(p1!=A.end()){ if(m==(k-1)){ return *p1; } p1++; m++; } while(p2!=B.end()){ if(m==(k-1)){ return *p2; } p2++; m++; } return -1; } };
4. Median of Two Sorted Arrays
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原文地址：http://www.cnblogs.com/li-daphne/p/5612240.html
