Median of Two Sorted Arrays[快排]-优快云博客

本文介绍了一种在O(log(m+n))的时间复杂度内找到两个已排序数组中位数的方法。通过实例演示了如何使用递归查找第k小元素的技巧来高效解决问题。

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)).

Example 1:

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

The median is 2.0

Example 2:

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

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

public class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
       int total = nums1.length+nums2.length;
        if(total%2==0){
            return (findKth(total/2+1, nums1, nums2, 0, 0)+findKth(total/2, nums1, nums2, 0, 0))/2.0;
        }else{
            return findKth(total/2+1, nums1, nums2, 0, 0);
        }
    }

    public static int findKth(int k, int[] nums1, int[] nums2, int s1, int s2){
        if(s1>=nums1.length)
            return nums2[s2+k-1];

        if(s2>=nums2.length)
            return nums1[s1+k-1];

        if(k==1)
            return Math.min(nums1[s1], nums2[s2]);

        int m1 = s1+k/2-1;
        int m2 = s2+k/2-1;

        int mid1 = m1<nums1.length?nums1[m1]:Integer.MAX_VALUE;
        int mid2 = m2<nums2.length?nums2[m2]:Integer.MAX_VALUE;

        if(mid1<mid2){
            return findKth(k-k/2, nums1, nums2, m1+1, s2);
        }else{
            return findKth(k-k/2, nums1, nums2, s1, m2+1);
        }
    }
}