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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]]
Note:
2 <= arr.length <= 10^5
-10^6 <= arr[i] <= 10^6
解析
根据题意，就是找出 arr 中差值最小的数值对有哪些。直接遍历经过排序的 arr ，如果两个相邻的元素差值比 min_diatance 小，那就更新 min_diatance ，然后更新 result 为空列表，并将这两个相邻的数值对放进去，如果两个相邻的元素差值等于 min_diatance ，将这两个相邻的数值对追加到 result 中，遍历 arr 结束之后，即可得到结果 。
解答
class Solution(object):
def minimumAbsDifference(self, arr):
"""
:type arr: List[int]
:rtype: List[List[int]]
"""
arr.sort()
i = 1
min_diatance = float("inf")
result = []
while i < len(arr):
distance = arr[i] - arr[i - 1]
if distance < min_diatance:
min_diatance = distance
result = [[arr[i - 1], arr[i]]]
elif distance == min_diatance:
result.append([arr[i - 1], arr[i]])
i += 1
return result
运行结果
Runtime: 304 ms, faster than 73.97% of Python online submissions for Minimum Absolute Difference.
Memory Usage: 24.4 MB, less than 42.56% of Python online submissions for Minimum Absolute Difference.
解析
另外,可以用比较高级的 python 内置函数 zip 来解决这个问题，先将 arr 从小到大进行排序，然后使用 zip 将 arr 和 arr[1:] 进行两两的差值计算，并得到最小的差值 m ，然后再重新遍历 arr 和 arr[1:] 组合而成的 zip 链，将差值为 m 的数值对找出来放入列表中即可。
这里简单介绍下 zip 的功能。zip 函数可以将可迭代的对象作为参数，将对象中对应的元素打包成一个个元组，然后返回由这些元组组成的列表。如果各个迭代器的元素个数不一致，则返回列表长度与最短的对象相同，利用 * 号操作符，可以将元组解压为列表。举例：
a = [1,2,3]
b = [4,5,6]
c = [4,5,6,7,8]
# 打包为元组的列表
print([x for x in zip(a,b)])
# 元素个数与最短的列表一致
print([x for x in zip(a,c)])
# # 与 zip 相反，*zipped 可理解为解压，返回二维矩阵式
print([x for x in zip(*zip(a,b))])
打印：
[(1, 4), (2, 5), (3, 6)]
[(1, 4), (2, 5), (3, 6)]
[(1, 2, 3), (4, 5, 6)]
解答
class Solution(object):
def minimumAbsDifference(self, arr):
"""
:type arr: List[int]
:rtype: List[List[int]]
"""
arr.sort()
m = min(j-i for i,j in zip(arr,arr[1:]))
return [[i,j] for i,j in zip(arr,arr[1:]) if j-i==m]
运行结果
Runtime: 340 ms, faster than 44.24% of Python online submissions for Minimum Absolute Difference.
Memory Usage: 32.2 MB, less than 14.55% of Python online submissions for Minimum Absolute Difference.
原题链接：leetcode.com/problems/mi…
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