[LeetCode] Median of Two Sorted Arrays

最新推荐文章于 2025-09-10 22:03:14 发布

原创最新推荐文章于 2025-09-10 22:03:14 发布 · 493 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #binary search

本文介绍了一种使用二分搜索递归的方法来查找两个已排序数组中的第K大元素，通过比较两个数组中相应位置的元素来缩小搜索范围，最终达到O(log(m+n))的时间复杂度。

This problem equals to finding the Kth maximum element in two sorted arrays. In order to reach the complexity of O(log(m+n)), we can use binary search recursively.

Compare the K/2th elements of A and B,

If A[k/2]<B[k/2], throw away the smaller part of A. Notice this won't affect the Kth element!

If there's less than k/2 elements in array, simply throw away this array. And the Kth element must be in another array!

class Solution {
public:
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        int s = m+n;
        if(s%2) return findKth(A,m,B,n,s/2+1);
        else return 0.5*(findKth(A,m,B,n,s/2)+findKth(A,m,B,n,s/2+1));
    }
    int findKth(int A[], int m, int B[], int n, int k) {
        if(m>n) return findKth(B,n,A,m,k);
        if(m==0) return B[k-1];
        if(k==1) return min(A[0],B[0]);
        
        int a = min(m,k/2);
        int b = k-a;
        if(A[a-1]<B[b-1]) return findKth(A+a,m-a,B,n,k-a);
        if(A[a-1]>B[b-1]) return findKth(A,m,B+b,n-b,k-b);
        return A[a-1];
    }
};