文章--LeetCode算法--MedianofTwoSortedArrays

目录

• MedianofTwoSortedArrays
  • 问题描述
  • 实例
  • 解决思路
  • 实现代码

MedianofTwoSortedArrays

问题描述

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)).

实例

• 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

解决思路

通过递归的方式不断趋近于中间值

实现代码

package com.leetcode.play;

/**
 * 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)).
 * <p>
 * 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 MedianofTwoSortedArrays {
    public static void main(String[] args) {
        int[] num1 = {1, 3};
        int[] num2 = {2};
        double i = findMedianSortedArrays(num1, num2);
        System.out.println(i);
    }

    public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        int l = (m + n + 1) >> 1;
        int r = (m + n + 2) >> 1;
        return (getkth(nums1, 0, nums2, 0, l) + getkth(nums1, 0, nums2, 0, r)) / 2.0;
    }

    public static double getkth(int[] A, int aStart, int[] B, int bStart, int k) {
        if (aStart == A.length)
            return B[bStart + k - 1];
        if (bStart == B.length)
            return A[aStart + k - 1];
        if (k == 1)
            return Math.min(A[aStart], B[bStart]);
        int aMid = Integer.MAX_VALUE, bMid = Integer.MAX_VALUE;
        if (aStart + k / 2 - 1 < A.length)
            aMid = A[aStart + k / 2 - 1];
        if (bStart + k / 2 - 1 < B.length)
            bMid = B[bStart + k / 2 - 1];
        if (aMid < bMid)
            return getkth(A, aStart + k / 2, B, bStart, k - k / 2);
        else
            return getkth(A, aStart, B, bStart + k / 2, k - k / 2);
    }
}