LeetCode:Median of Two Sorted Arrays(update)
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)).
解法一：合并
1 class Solution { 2 public: 3 double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { 4 vector<int> nums3; 5 int m=nums1.size(); 6 int n=nums2.size(); 7 nums3.reserve(m+n); 8 nums3.insert(nums3.end(),nums1.begin(),nums1.end()); 9 nums3.insert(nums3.end(),nums2.begin(),nums2.end()); 10 11 sort(nums3.begin(),nums3.end()); 12 if(nums3.size()%2) 13 return nums3[nums3.size()/2]; 14 else 15 return (double)(nums3[nums3.size()/2 - 1] + nums3[nums3.size()/2])/2; 16 17 }
解法二:来自官方的解题报告（）
1 class Solution { 2 public: 3 double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { 4 int m = nums1.size(), n = nums2.size(); 5 if (m > n) return findMedianSortedArrays(nums2, nums1); 6 int i, j, imin = 0, imax = m, half = (m + n + 1) / 2; 7 while (imin <= imax) { 8 i = (imin & imax) + ((imin ^ imax) >> 1); 9 j = half - i; 10 if (i > 0 && j < n && nums1[i - 1] > nums2[j]) imax = i - 1; 11 else if (j > 0 && i < m && nums2[j - 1] > nums1[i]) imin = i + 1; 12 else break; 13 } 14 int num1; 15 if (!i) num1 = nums2[j - 1]; 16 else if (!j) num1 = nums1[i - 1]; 17 else num1 = max(nums1[i - 1], nums2[j - 1]); 18 if ((m + n) & 1) return num1; 19 int num2; 20 if (i == m) num2 = nums2[j]; 21 else if (j == n) num2 = nums1[i]; 22 else num2 = min(nums1[i], nums2[j]); 23 return (num1 + num2) / 2.0; 24 } 25 };
