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This link has a very concise and fast solution based on binary search. Spend some time reading it and make sure you understand it. It is very helpful.
Since the C++ interface of this problem has been updated, I rewrite the code below.
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, num1, num2; 7 while (iMin <= iMax) { 8 i = (iMin + iMax) / 2; 9 j = half - i; 10 if (j > 0 && i < m && nums2[j - 1] > nums1[i]) 11 iMin = i + 1; 12 else if (i > 0 && j < n && nums1[i - 1] > nums2[j]) 13 iMax = i - 1; 14 else { 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 break; 19 } 20 } 21 if ((m + n) & 1) return num1; 22 if (i == m) num2 = nums2[j]; 23 else if (j == n) num2 = nums1[i]; 24 else num2 = min(nums1[i], nums2[j]); 25 return (num1 + num2) / 2.0; 26 } 27 };
[LeetCode] Median of Two Sorted Arrays
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原文地址:http://www.cnblogs.com/jcliBlogger/p/4619941.html