LeetCode004-Median of Two Sorted Arrays

Median of Two Sorted Arrays

Question:

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.

Example1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

解法代码

public class LeetCode004 {

    public static void main(String[] args) {
        int[] nums1 = new int[]{1,3};
        int[] nums2 = new int[]{2};
        
        double medianDouble = findMedianSortedArrays(nums1, nums2);
        System.out.println(medianDouble);
        
        int[] nums3 = new int[]{1,3};
        int[] nums4 = new int[]{2,4};
        
        medianDouble = findMedianSortedArrays(nums3, nums4);
        System.out.println(medianDouble);
        
        int[] nums5 = new int[]{1};
        int[] nums6 = new int[]{2};
        
        medianDouble = findMedianSortedArrays(nums5, nums6);
        System.out.println(medianDouble);
    }
    
    private static double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int[] numsTotal = new int[nums1.length + nums2.length];
        int i = 0,j = 0;
        
        while(i < nums1.length || j < nums2.length) {
            if(i < nums1.length && j < nums2.length) {
                if(nums1[i] > nums2[j]) {
                    numsTotal[i + j] = nums2[j];
                    j++;
                } else {
                    numsTotal[i + j] = nums1[i];
                    i++;
                }
            } else if(i == nums1.length) {
                numsTotal[i + j] = nums2[j];
                j++;
            } else if(j == nums2.length) {
                numsTotal[i + j] = nums1[i];
                i++;
            }
        }
        double medianDouble = (numsTotal.length % 2 == 0) ? 
                ((numsTotal[numsTotal.length/2 - 1]
                        + numsTotal[numsTotal.length/2])/2.0)
                : numsTotal[numsTotal.length/2];
        
        return medianDouble;
    }
}

Output：

2.0
2.5
1.5

Time And Space Complexity

• addTwoNumbers

Time: 需要循环两个数组，时间复杂度
Space: 新数组的长度为(m+n)

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