4. Median of Two Sorted Arrays

https://leetcode.com/problems/median-of-two-sorted-arrays/

C++

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size();
        int n = nums2.size();
        int mid = (m + n) / 2;
        int i = 0, j = 0;
        int prev = 0, next = 0;

        while ((i + j) <= mid) {
            prev = next;
            if (i >= m) {
                next = nums2[j++];
            } else if (j >= n) {
                next = nums1[i++];
            } else if (nums1[i] < nums2[j]) {
                next = nums1[i++];
            } else {
                next = nums2[j++];
            }
        }

        return (m + n) % 2 == 0 ? (prev + next) / 2.0 : next;
    }
};

Go

func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
  m, n := len(nums1), len(nums2)
  mid := (m + n) / 2
  i, j := 0, 0
  prev, next := 0, 0

  for i+j <= mid {
    prev = next

    if i >= m {
      next = nums2[j]
      j++
    } else if j >= n {
      next = nums1[i]
      i++
    } else if nums1[i] < nums2[j] {
      next = nums1[i]
      i++
    } else {
      next = nums2[j]
      j++
    }
  }

  if (m+n)%2 == 0 {
    return float64(prev+next) / 2
  } else {
    return float64(next)
  }
}

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {
  var m = nums1.length
  var n = nums2.length
  var mid = (m + n) / 2
  var i = 0
  var j = 0
  var prev = 0
  var next = 0

  while ((i + j) <= mid) {
    prev = next
    if (i >= m) {
      next = nums2[j++]
    } else if (j >= n) {
      next = nums1[i++]
    } else if (nums1[i] < nums2[j]) {
      next = nums1[i++]
    } else {
      next = nums2[j++]
    }
  }

  return (m + n) % 2 === 0 ? (prev + next) / 2 : next
}

Python

class Solution(object):

    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        m, n = len(nums1), len(nums2)
        mid = (m + n) / 2.0
        i, j = 0, 0
        prev, nxt = 0, 0

        while (i + j) <= mid:
            prev = nxt
            if i >= m:
                nxt = nums2[j]
                j += 1
            elif j >= n:
                nxt = nums1[i]
                i += 1
            elif nums1[i] < nums2[j]:
                nxt = nums1[i]
                i += 1
            else:
                nxt = nums2[j]
                j += 1

        return (prev + nxt) / 2.0 if (m + n) % 2 == 0 else nxt