【LeetCode】1200. Minimum Absolute Difference 解题报告 (C++)

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负雪明烛 2022-02-10 15:31:50 博主文章分类：LeetCode ©著作权

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作者：负雪明烛
id： fuxuemingzhu
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题目描述

Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements.

Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows

a, b are from arr
a < b
b - a equals to the minimum absolute difference of any two elements in arr

Example 1:

Input: arr = [4,2,1,3]
Output: [[1,2],[2,3],[3,4]]
Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.

Example 2:

Input: arr = [1,3,6,10,15]
Output: [[1,3]]

Example 3:

Input: arr = [3,8,-10,23,19,-4,-14,27]
Output: [[-14,-10],[19,23],[23,27]]

Constraints:

2 <= arr.length <= 10^5
-10^6 <= arr[i] <= 10^6

题目大意

给出了一个由不同数字构成的数组，哪些数字的差等于所有数字之差的最小值。

解题方法

排序

这个题肯定要先求所有数字差的最小值，暴力算是O(N^2)不可取。我们知道两个数字差最小，肯定是这两个数字比较接近，所以我们可以先排序，然后找到相邻数字的差的最小值。

找出所有数字差的最小值之后，再遍历一次，找出哪些相邻的数字差等于该最小值就行了。

C++代码如下：

class Solution {
public:
    vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
        const int N = arr.size();
        sort(arr.begin(), arr.end());
        int min_diff = INT_MAX;
        for (int i = 0; i < N - 1; ++i) {
            min_diff = min(min_diff, arr[i + 1] - arr[i]);
        }
        vector<vector<int>> res;
        for (int i = 0; i < N - 1; ++i) {
            if (arr[i + 1] - arr[i] == min_diff) {
                res.push_back({arr[i], arr[i + 1]});
            }
        }
        return res;
    }
};