[LeetCode] 910. Smallest Range II 最小区间之二

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).

After this process, we have some array?B.

Return the smallest possible difference between the maximum value of?B?and the minimum value of?B.

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2
Output: 6 Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3
Output: 3
Explanation: B = [4,6,3]

Note:

1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000

Github 同步地址:

https://github.com/grandyang/leetcode/issues/910

类似题目：

Smallest Range I

参考资料：

https://leetcode.com/problems/smallest-range-ii/

LeetCode All in One 题目讲解汇总(持续更新中...)

原文地址：https://www.cnblogs.com/grandyang/p/11361245.html

时间： 2024-10-09 12:16:51

[LeetCode] 910. Smallest Range II 最小区间之二的相关文章

LeetCode 910. Smallest Range II

很有意思的一道数学推理题目, 剪枝以后解法也很简洁.初看貌似需要把每个数跟其他数作比较.但排序以后可以发现情况大大简化:对于任一对元素a[i] < a[j], a[i] - k和a[j] + k 的情况可以排除, 因为会产生比原值更大的差, 所以对于原有数组的最小值min最大值max, (min - k, max + k)的情况可以排除.剩下的三种情况, (min - k, max - k), (min + k, max + k) 和 (min + k, max - k),后两种等价于原值max

[LeetCode] 632. Smallest Range Covering Elements from K Lists

[LeetCode]632. Smallest Range Covering Elements from K Lists 你有 k 个升序排列的整数数组.找到一个最小区间,使得 k 个列表中的每个列表至少有一个数包含在其中. 我们定义如果 b-a < d-c 或者在 b-a == d-c 时 a < c,则区间 [a,b] 比 [c,d] 小. 示例 1: 输入:[[4,10,15,24,26], [0,9,12,20], [5,18,22,30]] 输出: [20,24] 解释: 列表 1:

Smallest Range II

2020-01-21 21:43:52 问题描述: 问题求解: 这个题目还是有点难度的,感觉很巧妙也很难想到. 整体的思路如下: 1. 首先原问题等价于 +0 / + 2*K 2. 那么res = Max - Min 3. 不断更新Max,Min期望得到更小的res public int smallestRangeII(int[] A, int K) { int n = A.length; Arrays.sort(A); int max = A[0]; int min = A[0]; for (

[LeetCode] Trapping Rain Water II 收集雨水之二

Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining. Note: Both m and n are less than 110. The height of each unit cell is greater th

LeetCode 229. Majority Element II （众数之二）

Given an integer array of size n, find all elements that appear more than ? n/3 ? times. The algorithm should run in linear time and in O(1) space. 题目标签:Array 题目给了我们一个 nums array, 让我们找出所有数量超过 n/3 的众数.这一题与众数之一的区别在于,众数之一只要找到一个众数大于 n/2 的就可以.这一题要找到所

[LeetCode] 850. Rectangle Area II 矩形面积之二

We are given a list of (axis-aligned)?rectangles.? Each?rectangle[i] = [x1, y1, x2, y2]?, where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the?ith rectangle. Find the total area

[LeetCode] Reverse Linked List II 倒置链表之二

Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note:Given m, n satisfy the following condition:1 ≤ m ≤ n ≤ lengt

LeetCode 59 Spiral Matrix II 螺旋矩阵之二

Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. For example,Given n = 3, You should return the following matrix: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ] 此题跟之前那道本质上没什么区别,就相当于个类似逆运算的过程,这道题是要按螺旋的顺序来填数,由于

Leetcode 632.最小区间

最小区间你有 k 个升序排列的整数数组.找到一个最小区间,使得 k 个列表中的每个列表至少有一个数包含在其中. 我们定义如果 b-a < d-c 或者在 b-a == d-c 时 a < c,则区间 [a,b] 比 [c,d] 小. 示例 1: 输入:[[4,10,15,24,26], [0,9,12,20], [5,18,22,30]] 输出: [20,24] 解释: 列表 1:[4, 10, 15, 24, 26],24 在区间 [20,24] 中. 列表 2:[0, 9, 12, 20