4. Median of Two Sorted Arrays
- Total Accepted: 104147
- Total Submissions: 539044
- Difficulty: Hard
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 其实整个思路应该就是将两个有序数组进行排序的过程,然后根据总个数的奇偶决定中位数是一个还是两个的平均值。
1 public class No_4 { 2 public double findMedianSortedArrays(int[] nums1, int[] nums2) { 3 int total = nums1.length + nums2.length ; 4 int [] res ; 5 //判断最后的结果是单独一个数还是两个数的平均值 6 if(total%2 == 1){ 7 res = new int [1] ; 8 }else{ 9 res = new int [2] ; 10 } 11 int count = 0 ; //当前值是第count小的数 12 int i = 0 ; //nums1的下标 13 int j = 0 ; //nums2的下标 14 int k = 0 ; //res的下标 15 int min ; 16 /* 17 * 将两个有序数组进行排序的过程 18 * 当total为偶数时,结果为总数的第 (total-1)/2+1个数和第(total-1)/2+2个数 19 * 当total为基数时,结果为总数的第(total-1)/2+1个数 20 * 所以,当count > (total-1)/2 时,开始存储,直到res存满为止 21 */ 22 while((i < nums1.length ) && (j < nums2.length) && (k < res.length)){ 23 count++ ; 24 min = nums1[i] > nums2[j] ? nums2[j++] : nums1[i++] ; 25 if(count > (total-1)/2){ 26 res[k++] = min ; 27 } 28 } 29 //考虑nums2先排除完而没有完全得到需要的结果的情况 30 while((i < nums1.length ) && (k < res.length)){ 31 count++ ; 32 min = nums1[i++] ; 33 if(count > (total-1)/2){ 34 res[k++] = min ; 35 } 36 } 37 //考虑nums1先排除完而没有完全得到需要的结果的情况 38 while((j < nums2.length ) && (k < res.length)){ 39 count++ ; 40 min = nums2[j++] ; 41 if(count > (total-1)/2){ 42 res[k++] = min ; 43 } 44 } 45 46 //返回结果 47 if(total%2 == 1){ 48 return res[0] ; 49 }else{ 50 return ((res[0] + res[1])/2.0) ; 51 } 52 } 53 }
时间: 2024-07-30 13:36:25