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4. Median of Two Sorted Arrays

中文文档

Description

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

The overall run time complexity should be O(log (m+n)).

 

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.

 

Constraints:

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

Solutions

Binary search.

Python3

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        def findKth(i, j, k):
            if i >= m:
                return nums2[j + k - 1]
            if j >= n:
                return nums1[i + k - 1]
            if k == 1:
                return min(nums1[i], nums2[j])
            midVal1 = nums1[i + k // 2 - 1] if i + k // 2 - 1 < m else inf
            midVal2 = nums2[j + k // 2 - 1] if j + k // 2 - 1 < n else inf
            if midVal1 < midVal2:
                return findKth(i + k // 2, j, k - k // 2)
            return findKth(i, j + k // 2, k - k // 2)

        m, n = len(nums1), len(nums2)
        left, right = (m + n + 1) // 2, (m + n + 2) // 2
        return (findKth(0, 0, left) + findKth(0, 0, right)) / 2

Java

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        int left = (m + n + 1) / 2;
        int right = (m + n + 2) / 2;
        return (findKth(nums1, 0, nums2, 0, left) + findKth(nums1, 0, nums2, 0, right)) / 2.0;
    }

    private int findKth(int[] nums1, int i, int[] nums2, int j, int k) {
        if (i >= nums1.length) {
            return nums2[j + k - 1];
        }
        if (j >= nums2.length) {
            return nums1[i + k - 1];
        }
        if (k == 1) {
            return Math.min(nums1[i], nums2[j]);
        }
        int midVal1 = (i + k / 2 - 1 < nums1.length) ? nums1[i + k / 2 - 1] : Integer.MAX_VALUE;
        int midVal2 = (j + k / 2 - 1 < nums2.length) ? nums2[j + k / 2 - 1] : Integer.MAX_VALUE;
        if (midVal1 < midVal2) {
            return findKth(nums1, i + k / 2, nums2, j, k - k / 2);
        }
        return findKth(nums1, i, nums2, j + k / 2, k - k / 2);
    }
}

C++

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size();
        int n = nums2.size();
        int left = (m + n + 1) / 2;
        int right = (m + n + 2) / 2;
        return (findKth(nums1, 0, nums2, 0, left) + findKth(nums1, 0, nums2, 0, right)) / 2.0;
    }

    int findKth(vector<int>& nums1, int i, vector<int>& nums2, int j, int k) {
        if (i >= nums1.size()) return nums2[j + k - 1];
        if (j >= nums2.size()) return nums1[i + k - 1];
        if (k == 1) return min(nums1[i], nums2[j]);
        int midVal1 = i + k / 2 - 1 < nums1.size() ? nums1[i + k / 2 - 1] : INT_MAX;
        int midVal2 = j + k / 2 - 1 < nums2.size() ? nums2[j + k / 2 - 1] : INT_MAX;
        if (midVal1 < midVal2) return findKth(nums1, i + k / 2, nums2, j, k - k / 2);
        return findKth(nums1, i, nums2, j + k / 2, k - k / 2);
    }
};

Go

func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
	m, n := len(nums1), len(nums2)
	left, right := (m+n+1)/2, (m+n+2)/2
	var findKth func(i, j, k int) int
	findKth = func(i, j, k int) int {
		if i >= m {
			return nums2[j+k-1]
		}
		if j >= n {
			return nums1[i+k-1]
		}
		if k == 1 {
			return min(nums1[i], nums2[j])
		}
		midVal1 := math.MaxInt32
		midVal2 := math.MaxInt32
		if i+k/2-1 < m {
			midVal1 = nums1[i+k/2-1]
		}
		if j+k/2-1 < n {
			midVal2 = nums2[j+k/2-1]
		}
		if midVal1 < midVal2 {
			return findKth(i+k/2, j, k-k/2)
		}
		return findKth(i, j+k/2, k-k/2)
	}
	return (float64(findKth(0, 0, left)) + float64(findKth(0, 0, right))) / 2.0
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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