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 mini…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/smallest-range-ii/description/ 题目描述 Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add…

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…

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 mini…

题目如下: 解题思路:我的思路是先找出最大值.对于数组中任意一个元素A[i]来说,如果A[i] + K 是B中的最大值,那么意味着从A[i+1]开始的元素都要减去K,即如果有A[i] + K >= A[-1] - K,那么A[i] + K 就可以作为最大值而存在:如果A[i] + K是最大值,那么最大的最小值是多少呢?因为A[i] 左边的元素都比A[i]小,所以其左边的元素都可以加上K,最大的最小值就会在 A[0] + K 和 A[i+1] - K 之间产生.遍历数组,计算每一个A[i] + K…

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 mini…

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 (…

Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i]. 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…

You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the k lists. We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c. Example 1: Input:[…

You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the k lists. We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c. Example 1: Input:[…

You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the k lists. We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c. Example 1: Input:[…

You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the k lists. We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c. Example 1: Input:[…

You have k lists of sorted integers. Find the smallest range that includes at least one number from each of the k lists. For example, List 1: [4, 10, 15, 24, 26] List 2: [0, 9, 12, 20] List 3: [5, 18, 22, 30] The smallest range here would be [20, 24]…

You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the k lists. We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c. Example 1: Input:[…

problem 908. Smallest Range I solution: class Solution { public: int smallestRangeI(vector<int>& A, int K) { ], mn = A[]; for(auto a:A) { mx = max(mx, a); mn = min(mn, a); } , mx-mn-*K); } }; 参考 1. Leetcode_easy_908. Smallest Range I; 2. discuss…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/smallest-range/description/ 题目描述: You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each…

http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_3_B For a given array a1,a2,a3,...,aNa1,a2,a3,...,aN of NN elements and an integer KK, find the smallest sub-array size (smallest window length) where the elements in the sub-array contains…

Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i]. 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…

题目要求 Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i]. After this process, we have some array B. Return the smallest possible difference between the maximum value of B and the minimum value…

Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add xto A[i]. 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.…

题目 : Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i]. After this process, we have some array B. Return the smallest possible difference between the maximum value of B and the minimum value…

Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i]. 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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 数学计算 日期 题目地址:https://leetcode.com/problems/smallest-range-i/description/ 题目描述 Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and ad…

数据范围是3500,3500也就是说n的平方是可以接受的.这里告诉你就是有序的,也就是在提醒你可能会是一个类似于二分的算法,所以的话其实基于这两个认识的话我们就可以利用一个枚举叫二分的算法来解决这道题.怎么做呢?就首先的话我们要枚举一端,一端的话我们可以把所有的这个lists的里面所有元素都给去重然后对它进行左端的一个枚举,然后右端的话的呢我们就分别对每一个list进行一个二分来查找它可能包含所有至少来查找这个至少包含一个元素的最大的这么一个区间是什么.然后我们每次枚举的时候把这些最大的区间找一…

这是悦乐书的第348次更新,第372篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第213题(顺位题号是908).给定一个整数数组A,对于每个整数A[i],我们可以选择任何x,其中-K <= x <= K,并将x的值加到A[i]上.在这个过程之后,A变成了新数组B. 返回B的最大值和B的最小值之间的最小可能差值.例如: 输入:A = [1],K = 0 输出:0 说明:B = [1] 输入:A = [0,10],K = 2 输出:6 说明:B = [2,8] 输入…

题目如下: 解题思路:简单的不能再简单的题目了,对于任意一个A[i]来说,其可能的最小的最大值是A[i]-K,最大的最小值是A[i]+K.遍历数组,求出所有元素中最大的最小值和最小的最大值,两者之差(小于零则取零)就是答案. 代码如下: class Solution(object): def smallestRangeI(self, A, K): """ :type A: List[int] :type K: int :rtype: int """…

给定一个整数数组 A,对于每个整数 A[i],我们可以选择任意 x 满足 -K <= x <= K,并将 x 加到 A[i] 中. 在此过程之后,我们得到一些数组 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. 示例 1: 输入:A = [1], K = 0 输出:0 解释:B = [1] 示例 2: 输入:A = [0,10], K = 2 输出:6 解释:B = [2,8] 示例 3: 输入:A = [1,3,6], K = 3 输出:0 解释:B = [3,3,3] 或…

贪心基础 贪心(Greedy)常用于解决最优问题,以期通过某种策略获得一系列局部最优解.从而求得整体最优解. 贪心从局部最优角度考虑,只适用于具备无后效性的问题,即某个状态以前的过程不影响以后的状态.紧接下来的状态仅与当前状态有关.和分治.动态规划一样,贪心是一种思路,不是解决某类问题的具体方法. 应用贪心的关键,是甄别问题是否具备无后效性.找到获得局部最优的策略.有的问题比较浅显,例如一道找零钱的题目 LeetCode 860. Lemonade Change: // 860. Lemonad…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

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