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
Solution1: time complexity is O(m+n). merge two sorted array, then find the median
code:
1 /* 2 Time Complexity: O(m+n) 3 Space Complexity: O(m+n) 4 */ 5 6 class Solution { 7 public double findMedianSortedArrays(int[] nums1, int[] nums2) { 8 int[] mergedArray = mergeTwoSortedArray(nums1, nums2); 9 int n = mergedArray.length; 10 if( n % 2 == 0){ 11 return (mergedArray[(n-1)/2] + mergedArray[n/2])/2.0; 12 }else{ 13 return mergedArray[n/2]; 14 } 15 } 16 17 public int[] mergeTwoSortedArray(int[] nums1, int[] nums2){ 18 int[] merged = new int[nums1.length + nums2.length]; 19 int i = 0, j = 0, k = 0; 20 // 两个array同时扫 21 while( i < nums1.length && j < nums2.length){ 22 merged[k++] = (nums1[i] < nums2[j]) ? nums1[i++] : nums2[j++]; 23 } 24 //扫到只剩下较长的那个array， either nums1 or nums2 25 while ( i < nums1.length ){ 26 merged[k++] = nums1[i++]; 27 } 28 while(j < nums2.length){ 29 merged[k++] = nums2[j++]; 30 } 31 return merged; 32 } 33 }
Solution2: time complexity is O(log (m+n)).
[leetcode]4. Median of Two Sorted Arrays俩有序数组的中位数
原文：https://www.cnblogs.com/liuliu5151/p/10648755.html