描述 你是一个专业的小偷,计划偷窃沿街的房屋,每间房内都藏有一定的现金.这个地方所有的房屋都围成一圈,这意味着第一个房屋和最后一个房屋是紧挨着的.同时,相邻的房屋装有相互连通的防盗系统,如果两间相邻的房屋在同一晚上被小偷闯入,系统会自动报警. 给定一个代表每个房屋存放金额的非负整数数组,计算你在不触动警报装置的情况下,能够偷窃到的最高金额. 示例 1: 输入: [2,3,2]输出: 3解释: 你不能先偷窃 1 号房屋(金额 = 2),然后偷窃 3 号房屋(金额 = 2), 因为他们是相邻的.示例…

https://leetcode.com/problems/house-robber/ 题意: 一维数组,相加不相邻的数组,返回最大的结果. 思路: 一开始思路就是DP,用一维数组保存dp[i]保存如果偷第i间,此时可偷到多少.DP的方向不太好,所以效率很低. Runtime: 4 ms, faster than 17.53% class Solution { public: int rob(vector<int> &nums) { ; int len = nums.size(); )…

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will autom…

leetcode:234 回文链表 关键点:请判断一个链表是否为回文链表.示例 1:输入: 1->2输出: false示例 2:输入: 1->2->2->1输出: true. 1/** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 *     int val; 5 *     ListNode *next; 6 *     ListNode(int x) : val(x), next(NULL) {}…

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will autom…

198. 打家劫舍 198. House Robber 题目描述 你是一个专业的小偷,计划偷窃沿街的房屋.每间房内都藏有一定的现金,影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统,如果两间相邻的房屋在同一晚上被小偷闯入,系统会自动报警. 给定一个代表每个房屋存放金额的非负整数数组,计算你在不触动警报装置的情况下,能够偷窃到的最高金额. 每日一算法2019/5/8Day 5LeetCode198. House Robber 示例 1: 输入: [1,2,3,1] 输出: 4 解释:…

198. House Robber You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected…

讲解视频见刘宇波leetcode动态规划第三个视频 记忆化搜索代码: #include <bits/stdc++.h> using namespace std; class Solution { private: vector<int>memo; int tryRob(int index, vector<int>& nums) { if (index > nums.size()) { ; } ) return memo[index]; int i; ; f…

234. Palindrome Linked List 1. 使用快慢指针找中点的原理是fast和slow两个指针,每次快指针走两步,慢指针走一步,等快指针走完时,慢指针的位置就是中点.如果是偶数个数,正好是一半一半,如果是奇数个数,慢指针正好在中间位置,判断回文的时候不需要比较该位置数据. 注意好好理解快慢指针的算法原理及应用. 2. 每次慢指针走一步,都把值存入栈中,等到达中点时,链表的前半段都存入栈中了,由于栈的后进先出的性质,就可以和后半段链表按照回文对应的顺序比较. solution…

Implement atoi to convert a string to an integer. Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases. Notes: It is intended for this problem to be spe…