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力扣解法汇总1200-最小绝对差
2022-07-05 16:55:00 【失落夏天】
目录链接:
力扣编程题-解法汇总_分享+记录-CSDN博客
GitHub同步刷题项目:
https://github.com/September26/java-algorithms
原题链接:力扣
描述:
给你个整数数组 arr,其中每个元素都 不相同。
请你找到所有具有最小绝对差的元素对,并且按升序的顺序返回。
示例 1:
输入:arr = [4,2,1,3]
输出:[[1,2],[2,3],[3,4]]
示例 2:
输入:arr = [1,3,6,10,15]
输出:[[1,3]]
示例 3:
输入:arr = [3,8,-10,23,19,-4,-14,27]
输出:[[-14,-10],[19,23],[23,27]]
提示:
2 <= arr.length <= 10^5
-10^6 <= arr[i] <= 10^6
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/minimum-absolute-difference
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
解题思路:
* 解题思路: * 这题比较简单,排序之后,最小值一定是相邻的两者之间的最差值。
代码:
public class Solution1200 {
public List<List<Integer>> minimumAbsDifference(int[] arr) {
Arrays.sort(arr);
List<List<Integer>> result = new ArrayList<>();
int minDIff = Integer.MAX_VALUE;
for (int i = 1; i < arr.length; i++) {
int diffValue = arr[i] - arr[i - 1];
if (diffValue < minDIff) {
result.clear();
result.add(createList(arr[i], arr[i - 1]));
minDIff = diffValue;
continue;
}
if (diffValue == minDIff) {
result.add(createList(arr[i], arr[i - 1]));
continue;
}
}
return result;
}
private List<Integer> createList(int i1, int i2) {
ArrayList<Integer> integers = new ArrayList<>();
integers.add(i2);
integers.add(i1);
return integers;
}
}
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