如何找到2个排序的阵列A和B的中值,它们分别是长度为m和n。 我搜索过,但大多数算法都假设两个数组的大小相同。我想知道如果m!= n我们怎么能找到中位数 考虑例子, A = {1,3,5,7,11,15},其中m = 6, B = {2,4,8,12,14},其中n = 5 中位数为7
感谢任何帮助。 我正准备接受采访,现在我正在努力解决这个问题。
答案 0 :(得分:7)
这是用于查找两个不等长度的排序数组的中位数的JAVA代码
package FindMedianBetween2SortedArraysOfUnequalLength;
import java.util.Arrays;
import java.util.Scanner;
public class UsingKthSmallestElementLogic {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
try{
System.out.println("Enter the number of elements in the first SORTED array");
int n = in.nextInt();
int[] array1 = new int[n];
System.out.println("Enter the elements of the first SORTED array");
for(int i=0;i<n;i++)
array1[i]=in.nextInt();
System.out.println("Enter the number of elements in the second SORTED array");
int m = in.nextInt();
int[] array2 = new int[m];
System.out.println("Enter the elements of the second SORTED array");
for(int i=0;i<m;i++)
array2[i]=in.nextInt();
System.out.println("Median of the two SORTED arrays is: "+findMedian(array1,array2,array1.length,array2.length));
}
finally{
in.close();
}
}
private static int findMedian(int[] a, int[] b,
int aLength, int bLength) {
int left = (aLength+bLength+1)>>1;
int right = (aLength+bLength+2)>>1;
return ((findKthSmallestElement(a,b,a.length,b.length,left)+findKthSmallestElement(a,b,a.length,b.length,right))/2);
}
private static int findKthSmallestElement(int[] a, int[] b,
int aLength, int bLength, int k) { // All the 5 parameters passed are VERY VERY IMP
/* to maintain uniformity, we will assume that size_a is smaller than size_b
else we will swap array in call :) */
if(aLength>bLength)
return findKthSmallestElement(b, a, bLength, aLength, k);
/* We have TWO BASE CASES
* Now case when size of smaller array is 0 i.e there is no elemt in one array*/
//BASE CASE 1. If the smallest array length is 0
if(aLength == 0 && bLength > 0)
return b[k-1]; // due to zero based index
/* case where k==1 that means we have hit limit */
//BASE CASE 2. If k==1
if(k==1)
return Math.min(a[0], b[0]);
/* Now the divide and conquer part */
int i = Math.min(aLength, k/2) ; // k should be less than the size of array
int j = Math.min(bLength, k/2) ; // k should be less than the size of array
if(a[i-1] > b[j-1])
// Now we need to find only K-j th element
return findKthSmallestElement(a, Arrays.copyOfRange(b, j, b.length), a.length, b.length -j, k-j);
else
return findKthSmallestElement(Arrays.copyOfRange(a, i, a.length), b, a.length-i, b.length, k-i);
}
}
/*
Analysis:
Time Complexity = O(log(n+m))
Space Complexity = O(1)*/
答案 1 :(得分:2)
对中间有序元素的线性搜索将是具有恒定空间的O(m + n)。这不是最佳选择,但在面试中制作是切合实际的。
numElements = arr1.length + arr2.length
elementsToInspect = floor(numElements / 2)
i1 = 0
i2 = 0
if elementsToInspect == 0
return arr1[i1]
while (1) {
if (arr1[i1] < arr2[i2]) {
i1++
elementsToInspect--
if elementsToInspect == 0
return arr1[i1]
} else {
i2++
elementsToInspect--
if elementsToInspect == 0
return arr2[i2]
}
}
答案 2 :(得分:0)
A simple O(log(m + n)) solution: watch video https://www.youtube.com/watch?v=LPFhl65R7ww
class Solution {
public double findMedianSortedArrays(int[] arr1, int[] arr2) {
if(arr1 == null && arr2 == null) {
return -1;
}
if(arr1.length == 0 && arr2.length == 0) {
return -1;
}
int x = arr1.length;
int y = arr2.length;
if(x > y) {
return findMedianSortedArrays(arr2, arr1);
}
int low = 0;
int high = x;
while (low <= high) {
int partitionX = (low + high) / 2;
int partitionY = (x + y + 1) / 2 - partitionX;
int maxX = partitionX == 0 ? Integer.MIN_VALUE : arr1[partitionX - 1];
int minX = partitionX == x ? Integer.MAX_VALUE : arr1[partitionX];
int maxY = partitionY == 0 ? Integer.MIN_VALUE : arr2[partitionY - 1];
int minY = partitionY == y ? Integer.MAX_VALUE : arr2[partitionY];
if(maxX <= minY && maxY <= minX) {
if((x + y)%2 == 0) {
return (Math.max(maxX, maxY) + Math.min(minX, minY)) / 2.0;
} else {
return Math.max(maxX, maxY);
}
} else if(maxX > minY) {
high = partitionX - 1;
} else {
low = partitionX + 1;
}
}
return -1;
}
}