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LeetCode-Median of Two Sorted Arrays

题目


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:

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

The median is 2.0

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

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

代码


// solution 1
public class Solution {
public double FindMedianSortedArrays(int[] nums1, int[] nums2)
{
var m = nums1.Length;
var n = nums2.Length;
if (m > n)
{
(nums1, nums2) = (nums2, nums1);
// return FindMedianSortedArrays(nums2, nums1);
}
// 左侧元素个数
var lc = (m + n + 1) / 2;
// 在 nums1 的[0, m]区间寻找合适的分割位置
// 满足 nums1[i-1] <= nums2[j] && nums1[j-1] <= nums2[i]
var l = 0;
var r = m;
while (l < r)
{
var i = (l + r + 1) / 2;
var j = lc - i;
if (nums1[i - 1] > nums2[j])
{
r = i - 1; //[l, i-1]
}
else
{
l = i; //[i, r]
}
}
{
var i = l;
var j = lc - l;
var m1 = i == 0 ? int.MinValue : nums1[i - 1];
var n1 = j == 0 ? int.MinValue : nums2[j - 1];
var m2 = i == m ? int.MaxValue : nums1[i];
var n2 = j == n ? int.MaxValue : nums2[j];
if ((m + n) % 2 == 1)
{
return Math.Max(m1, n1);
}
return (Math.Max(m1, n1) + Math.Min(m2, n2)) / 2.0;
}
}
}