题目描述: 自己的解: class Solution: def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int: res = n = len(points) : return res pre = points[] for i,v in enumerate(points): : res += min(abs(v[]-pre[]),abs(v[]-pre[])) + abs(abs(v[]-pre[])-abs(v[]-pr…
Given an unsorted array, find the maximum difference between the successive elements in its sorted form. Return 0 if the array contains less than 2 elements. Example 1: Input: [3,6,9,1] Output: 3 Explanation: The sorted form of the array is [1,3,6,9]…
164. Maximum Gap 164. 最大间隔 Given an unsorted array, find the maximum difference between the successive elements in its sorted form. Return 0 if the array contains less than 2 elements. 给定一个未排序数组,找出该数组在有序形式时,连续元素之间最大的间隔. Example 1: Input: [3,6,9,1] Ou…
Given an unsorted array, find the maximum difference between the successive elements in its sorted form. Try to solve it in linear time/space. Return 0 if the array contains less than 2 elements. You may assume all elements in the array are non-negat…
Given an unsorted array, find the maximum difference between the successive elements in its sorted form. Try to solve it in linear time/space. Return 0 if the array contains less than 2 elements. You may assume all elements in the array are non-negat…
5629. 重新格式化电话号码 模拟 注意一些细节,最后位置是否取值. class Solution { public: string reformatNumber(string number) { string s; for (auto c: number) if (c != ' ' && c != '-') s += c; string res; for (int i = 0; i < s.size();) { if ((int)s.size() - i > 4) res…
Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 二叉树的经典问题之最小深度问题就是就最短路径的节点个数,还是用深度优先搜索DFS来完成,万能的递归啊...请看代码: /** * Definition for binary tre…
On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step w…
Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Your goal is to reach the last index in the minimum number of jumps…