文章--LeetCode算法--MedianofTwoSortedArrays
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);
}
}