#387 The Smallest Difference_387. the smallest difference-优快云博客

本文链接：https://blog.youkuaiyun.com/haifischxia/article/details/52189190

本文介绍了一种算法问题，即给定两个整数数组A和B，如何找到两数组中的元素A[i]和B[j]使得它们之间的绝对差值最小，并返回这个最小差值。文章提出了一种使用双指针技巧的解决方案，该方法首先对两个数组进行排序，然后通过二分查找的方法找到可能的最小差值。

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题目描述：

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.

Have you met this question in a real interview?

Yes

Example

For example, given array A = [3,6,7,4], B = [2,8,9,3], return 0

Challenge

O(n log n) time

题目思路：

这题最简单的思路就是遍历两个数组，找到最小diff。题目的要求是O(nlogn)的时间复杂度，那么就会想到two pointers的做法：

1. 首先，two pointers做法一般都用在有序数组上，所以上来先把A和B排序；

2. 然后，将A中的element A[i]一个一个拿出来，跟B里的比较。这里不需要将B中的每一个element都拿来比较，其实最需要比的只是与A[i]相近的那些。由此我们可以利用有序数组的特点，做一个二分搜索，挑最重要的那些来比。比如，如果A[i]比B[mid]大，那么diff有可能更小的就只能在B[mid]之后的那些数。这里的code我就偷懒了，每次求得的mid都做一个min，其实对于每一个A[i], 真正需要比较的是B中最后一个比A[i]小，以及第一个比A[i]大的数。

Mycode（AC = 929ms）:

class Solution {
public:
    /**
     * @param A, B: Two integer arrays.
     * @return: Their smallest difference.
     */
    int smallestDifference(vector<int> &A, vector<int> &B) {
        // write your code here
        sort(A.begin(), A.end()); // A = 3,4,6,7
        sort(B.begin(), B.end()); // B = 2,4,8,9
        
        int diff = INT_MAX;
        for (int i = 0; i < A.size(); i++) { // A[i] = 4
            int l = 0, r = B.size() - 1; // l = 0, r = 3
            
            while (l < r) {
                int mid = (l + r) / 2; // mid = 1
                
                if (A[i] < B[mid]) {
                    r = mid - 1;
                    diff = min(diff, abs(A[i] - B[mid]));
                }
                else if (A[i] > B[mid]) {
                    l = mid + 1;
                    diff = min(diff, abs(A[i] - B[mid]));
                }
                else {
                    return 0;
                }
            }
            
            if (l >= 0 && l < B.size()) {
                diff = min(diff, abs(A[i] - B[l])); // diff = 1
            }
        }
        
        return diff;
    }
};