[LeetCode] 4. Median of Two Sorted Arrays.（两个有序数组的中位数）

• Difficulty: Hard
• Related Topics: Array, Binary Search, Divide and Conquer
• Link: https://leetcode.com/problems/median-of-two-sorted-arrays/

Description

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
给定两有序数组 nums1 和 nums2，二者的大小分别为 m 和 n，返回这两组数的中位数。

Follow up

The overall run time complexity should be (O(log (m + n)))
总时间复杂度应为 (O(log (m + n)))

Examples

Example 1

Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.

Example 2

Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.

Example 3

Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000

Example 4

Input: nums1 = [], nums2 = [1]
Output: 1.00000

Example 5

Input: nums1 = [2], nums2 = []
Output: 2.00000

Constraints

• nums1.length == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= m + n <= 2000
• -106 <= nums1[i], nums2[i] <= 1e6

Solution

这题我也是看了题解才明白个大概。参考文章见后，代码如下：

import kotlin.math.max
import kotlin.math.min

class Solution {
    fun findMedianSortedArrays(nums1: IntArray, nums2: IntArray): Double {
        // 该算法需保证 nums2 更大
        if (nums1.size > nums2.size) {
            return findMedianSortedArrays(nums2, nums1)
        }

        var iMin = 0
        var iMax = nums1.size
        val halfLen = (nums1.size + nums2.size + 1) / 2

        while (iMin <= iMax) {
            val i = (iMin + iMax) / 2
            val j = halfLen - i
            if (i < nums1.size && nums2[j - 1] > nums1[i]) {
                iMin = i + 1
            } else if (i > 0 && nums1[i - 1] > nums2[j]) {
                iMax = i - 1
            } else {
                val maxOfLeft = when {
                    i == 0 -> nums2[j - 1]
                    j == 0 -> nums1[i - 1]
                    else -> max(nums1[i - 1], nums2[j - 1])
                }
                if ((nums1.size + nums2.size) % 2 == 1) {
                    return maxOfLeft.toDouble()
                }

                val minOfRight = when {
                    i == nums1.size -> nums2[j]
                    j == nums2.size -> nums1[i]
                    else -> min(nums1[i], nums2[j])
                }

                return (maxOfLeft + minOfRight) / 2.0
            }
        }
        
        return 0.0
    }
}

Reference

1. https://zhuanlan.zhihu.com/p/70654378