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82 lines
1.7 KiB
Markdown
82 lines
1.7 KiB
Markdown
# [910. Smallest Range II](https://leetcode.com/problems/smallest-range-ii/)
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## 题目
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Given an array `A` of integers, for each integer `A[i]` we need to choose **either `x = -K` or `x = K`**, and add `x` to `A[i]` **(only once)**.
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After this process, we have some array `B`.
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Return the smallest possible difference between the maximum value of `B` and the minimum value of `B`.
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**Example 1:**
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```c
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Input: A = [1], K = 0
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Output: 0
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Explanation: B = [1]
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```
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**Example 2:**
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```c
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Input: A = [0,10], K = 2
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Output: 6
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Explanation: B = [2,8]
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```
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**Example 3:**
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```c
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Input: A = [1,3,6], K = 3
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Output: 3
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Explanation: B = [4,6,3]
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```
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**Note:**
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1. `1 <= A.length <= 10000`
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2. `0 <= A[i] <= 10000`
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3. `0 <= K <= 10000`
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## 题目大意
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给你一个整数数组 A，对于每个整数 A[i]，可以选择 x = -K 或是 x = K （K 总是非负整数），并将 x 加到 A[i] 中。在此过程之后，得到数组 B。返回 B 的最大值和 B 的最小值之间可能存在的最小差值。
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## 解题思路
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- 简单题。先排序，找出 A 数组中最大的差值。然后循环扫一遍数组，利用双指针，选择 x = -K 或是 x = K ，每次选择都更新一次最大值和最小值之间的最小差值。循环一次以后便可以找到满足题意的答案。
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## 代码
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```go
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package leetcode
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import "sort"
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func smallestRangeII(A []int, K int) int {
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n := len(A)
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sort.Ints(A)
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res := A[n-1] - A[0]
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for i := 0; i < n-1; i++ {
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a, b := A[i], A[i+1]
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high := max(A[n-1]-K, a+K)
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low := min(A[0]+K, b-K)
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res = min(res, high-low)
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}
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return res
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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``` |