1200. Minimum Absolute Difference

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

这道题给了一个没有重复数字的整型数组,现在让找出差的绝对值最小的数对儿。既然是 Easy 的身价,那么就没有太 Fancy 的解法,为了更方便的找出差的绝对值最小的数对儿,先给数组进行排序,这样最小差值一定会出现在相邻的两个数字之间。接下来就是遍历所有相邻的两个数字,维护一个最小值 mn,若当前差值 diff 小于等于 mn,则进行进一步操作,二者中唯一不同的是当 diff 小于 mn 时,结果 res 需要清空。然后将 mn 更新为 diff,并把当前数组对儿加入到结果 res 中即可,参见代码如下:

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

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1200

参考资料:

https://leetcode.com/problems/minimum-absolute-difference/

https://leetcode.com/problems/minimum-absolute-difference/discuss/388289/Java-sorting-beats-100-explained

LeetCode All in One 题目讲解汇总(持续更新中…)


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