[LeetCode] 1200. Minimum Absolute Difference

Easy

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：将数组中两数间隔最小的组合成对输出。

例如数组[4,2,1,3]，最后结果是[[1,2],[2,3],[3,4]]。因为输出的每组数字两数只差都为1，且数组中没有哪两个数只差比1更小。

方法，为了尽快找到差值最小的组合，所以先对数组尽心排序，然后维护一个记录最小差值的数字。如果碰到有更小的差值就把结果清空，再将这个组合放入结果中，如果组合差值和当前的最小差值相等就把这个组合叶放入结果中。

代码如下：

class Solution {
public:
    vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
        sort(arr.begin(),arr.end());
        vector<vector<int>> res={{arr[0],arr[1]}};
        int diff=arr[1]-arr[0];
        for(int i=1;i<arr.size()-1;++i){
            if(arr[i+1]-arr[i]<diff){
                res.clear();
                diff = arr[i + 1] - arr[i];
            }
            else if(arr[i+1]-arr[i]>diff){
                continue;
            }
            res.push_back({arr[i],arr[i+1]});
        }
        return res;
    }
};