The Smallest Difference
Given two array of integers(the first array is array A
, the second array is array B
), now we are going to find a element in array A which is A[i], and another element in array B which is B[j], so that the difference between A[i] and B[j] (|A[i] - B[j]|) is as small as possible, return their smallest difference.
Example
For example, given array A = [3,6,7,4]
, B = [2,8,9,3]
, return 0.
解题思路:1. 最直接的做法是用两个for循环遍历两个数组,其时间复杂度为O(n2), 题目要求用O(nlogn)的时间复杂度来实现,因此可以考虑对数组进行排序。
2. 排序之后,两个数组都为升序数组。可以发现,内层的for循环不用每次都从0开始。遍历的时间复杂度变为O(n)。
public class Solution { /** * @param A, B: Two integer arrays. * @return: Their smallest difference. */ public int smallestDifference(int[] A, int[] B) { // write your code here if (A == null || B == null || A.length == 0 || B.length == 0) { return -1; } Arrays.sort(A); Arrays.sort(B); int indexA = 0, indexB = 0; int min = Integer.MAX_VALUE; while (indexA < A.length && indexB < B.length) { min = Math.min(min, Math.abs(A[indexA] - B[indexB])); if (A[indexA] < B[indexB]) { ++indexA; } else { ++indexB; } } return min; } }