The Smallest Difference

本文介绍了一种在两个整数数组中找到元素间最小绝对差的方法,并提供了一个示例代码实现,通过排序和双指针技巧高效解决问题。

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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

来一道简单的题

感觉最近遇到瓶颈了,不能够做到举一反三。感觉还是一些东西没有真正领会到。还需要不断地理解。


这道题要找两个数组中的最小差距。提示给了复杂度nlogn, 自然想到排序。然后两个指针一次增长,求距离就可以了。

代码:

public int smallestDifference(int[] A, int[] B) {
        // write your code here
        
        Arrays.sort(A);
        Arrays.sort(B);
        
        int index1 = 0;
        int index2 = 0;
        int min = Integer.MAX_VALUE;
        while(index1<A.length && index2<B.length){
             min = Math.min(min, Math.abs(A[index1] -  B[index2]));
             if(A[index1]>B[index2]){
                 index2++;
             }else if(A[index1]<B[index2]){
                 index1++;
             }else{
                 index1++;
                 index2++;
             }
        }
        return min;
    }
}


Load balancing (负载均衡) refers to efficiently distributing incoming network traffic across a group of backend servers. A load balancing algorithm distributes loads in a specific way. If we can estimate the maximum incoming traffic load, here is an algorithm that works according to the following rule: The incoming traffic load of size S will first be partitioned into two parts, and each part may be again partitioned into two parts, and so on. Only one partition is made at a time. At any time, the size of the smallest load must be strictly greater than half of the size of the largest load. All the sizes are positive integers. This partition process goes on until it is impossible to make any further partition. For example, if S=7, then we can break it into 3+4 first, then continue as 4=2+2. The process stops at requiring three servers, holding loads 3, 2, and 2. Your job is to decide the maximum number of backend servers required by this algorithm. Since such kind of partitions may not be unique, find the best solution -- that is, the difference between the largest and the smallest sizes is minimized. Input Specification: Each input file contains one test case, which gives a positive integer S (2≤N≤200), the size of the incoming traffic load. Output Specification: For each case, print two numbers in a line, namely, M, the maximum number of backend servers required, and D, the minimum of the difference between the largest and the smallest sizes in a partition with M servers. The numbers in a line must be separated by one space, and there must be no extra space at the beginning or the end of the line. Sample Input: 22 Sample Output: 4 1 Hint: There are more than one way to partition the load. For example: 22 = 8 + 14 = 8 + 7 + 7 = 4 + 4 + 7 + 7 or 22 = 10 + 12 = 10 + 6 + 6 = 4 + 6 + 6 + 6 or 22 = 10 + 12 = 10 + 6 + 6 = 5 + 5 + 6 + 6 All requires 4 servers. The last partition has the smallest difference 6−5=1, hence 1 is printed out.请给这一到题设置测试用例,包含最大最小情况,和一些特殊的情况,以及对于回溯算法最好和最坏的情况
最新发布
06-03
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