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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 f(i: int, j: int, 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])
            p = k // 2
            x = nums1[i + p - 1] if i + p - 1 < m else inf
            y = nums2[j + p - 1] if j + p - 1 < n else inf
            return f(i + p, j, k - p) if x < y else f(i, j + p, k - p)

        m, n = len(nums1), len(nums2)
        a = f(0, 0, (m + n + 1) // 2)
        b = f(0, 0, (m + n + 2) // 2)
        return (a + b) / 2

Java

class Solution {
    private int m;
    private int n;
    private int[] nums1;
    private int[] nums2;

    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        m = nums1.length;
        n = nums2.length;
        this.nums1 = nums1;
        this.nums2 = nums2;
        int a = f(0, 0, (m + n + 1) / 2);
        int b = f(0, 0, (m + n + 2) / 2);
        return (a + b) / 2.0;
    }

    private int f(int i, int j, int k) {
        if (i >= m) {
            return nums2[j + k - 1];
        }
        if (j >= n) {
            return nums1[i + k - 1];
        }
        if (k == 1) {
            return Math.min(nums1[i], nums2[j]);
        }
        int p = k / 2;
        int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
        int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
        return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
    }
}

C++

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        function<int(int, int, int)> f = [&](int i, int j, int 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]);
            }
            int p = k / 2;
            int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
            int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
            return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
        };
        int a = f(0, 0, (m + n + 1) / 2);
        int b = f(0, 0, (m + n + 2) / 2);
        return (a + b) / 2.0;
    }
};

Go

func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
	m, n := len(nums1), len(nums2)
	var f func(i, j, k int) int
	f = 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])
		}
		p := k / 2
		x, y := 1<<30, 1<<30
		if ni := i + p - 1; ni < m {
			x = nums1[ni]
		}
		if nj := j + p - 1; nj < n {
			y = nums2[nj]
		}
		if x < y {
			return f(i+p, j, k-p)
		}
		return f(i, j+p, k-p)
	}
	a, b := f(0, 0, (m+n+1)/2), f(0, 0, (m+n+2)/2)
	return float64(a+b) / 2.0
}

TypeScript

function findMedianSortedArrays(nums1: number[], nums2: number[]): number {
    const m = nums1.length;
    const n = nums2.length;
    const f = (i: number, j: number, k: number): number => {
        if (i >= m) {
            return nums2[j + k - 1];
        }
        if (j >= n) {
            return nums1[i + k - 1];
        }
        if (k == 1) {
            return Math.min(nums1[i], nums2[j]);
        }
        const p = Math.floor(k / 2);
        const x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
        const y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
        return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
    };
    const a = f(0, 0, Math.floor((m + n + 1) / 2));
    const b = f(0, 0, Math.floor((m + n + 2) / 2));
    return (a + b) / 2;
}

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function (nums1, nums2) {
    const m = nums1.length;
    const n = nums2.length;
    const f = (i, j, k) => {
        if (i >= m) {
            return nums2[j + k - 1];
        }
        if (j >= n) {
            return nums1[i + k - 1];
        }
        if (k == 1) {
            return Math.min(nums1[i], nums2[j]);
        }
        const p = Math.floor(k / 2);
        const x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
        const y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
        return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
    };
    const a = f(0, 0, Math.floor((m + n + 1) / 2));
    const b = f(0, 0, Math.floor((m + n + 2) / 2));
    return (a + b) / 2;
};

TypeScript

public class Solution {
    private int m;
    private int n;
    private int[] nums1;
    private int[] nums2;

    public double FindMedianSortedArrays(int[] nums1, int[] nums2) {
        m = nums1.Length;
        n = nums2.Length;
        this.nums1 = nums1;
        this.nums2 = nums2;
        int a = f(0, 0, (m + n + 1) / 2);
        int b = f(0, 0, (m + n + 2) / 2);
        return (a + b) / 2.0;
    }

    private int f(int i, int j, int k) {
        if (i >= m) {
            return nums2[j + k - 1];
        }
        if (j >= n) {
            return nums1[i + k - 1];
        }
        if (k == 1) {
            return Math.Min(nums1[i], nums2[j]);
        }
        int p = k / 2;
        int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
        int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
        return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
    }
}

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