House Robber:不能相邻,求能获得的最大值 House Robber II:不能相邻且第一个和最后一个不能同时取,求能获得的最大值 House Robber III:二叉树下的不能相邻,求能获得的最大值 Paint House:用3种颜色,相邻的房屋不能用同一种颜色,求花费最小 Paint House II:用k种颜色,相邻的房屋不能用同一种颜色,求花费最小Paint Fence:用k种颜色,相邻的可以用同一种颜色,但不能超过连续的2个,求有多少种可能性 198. House Robb…

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…

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all hou…

337. House Robber III Total Accepted: 18475 Total Submissions: 47725 Difficulty: Medium The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has…

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…

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…

打家劫舍 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. 示例 1: 输入: [1,2,3,1] 输出: 4 解释: 偷窃 1 号房屋 (金额 = 1) ,然后偷窃 3 号房屋 (金额 = 3).   偷窃到的最高金额 = 1 + 3 = 4 . 示例…

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:…

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…

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…

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…

题目描述: 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…

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…

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(); )…

题意是强盗能隔个马抢马,看如何获得的价值最高 动态规划题需要考虑状态,阶段,还有状态转移,这个可以参考<动态规划经典教程>,网上有的下的,里面有大量的经典题目讲解 dp[i]表示到第i匹马时的最大价值是多少, 因此所有的dp[i] = max(dp[i-2]+nums[i],dp[i-1]) (其中dp[0] = nums[0] dp[1] = max(nums[0],nums[1]): class Solution { public: int rob(vector<int>&am…

function rob(nums) { if(!nums || nums.length === 0) { return 0; } else if(nums.length < 2){ return nums[0]; } let memo = new Array(nums.length); memo[0] = nums[0]; memo[1] = Math.max(nums[0], nums[1]); for(let i = 2; i < nums.length; i++) { memo[i]…

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all hou…

题目描述: The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "a…

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…

Total Accepted: 1341 Total Submissions: 3744 Difficulty: Medium The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent…

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/house-robber-iii/description/ 题目描述 The thief has found himself a new place for his thievery again. There is only one entrance to this area…

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all hou…

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all hou…

每个节点是个房间,数值代表钱.小偷偷里面的钱,不能偷连续的房间,至少要隔一个.问最多能偷多少钱 TreeNode* cur mp[{cur, true}]表示以cur为根的树,最多能偷的钱 mp[{cur, false}]表示以cur为根的树,不偷cur节点的钱,最多能偷的钱 可以看出有下面的关系 mp[{node, false}] = mp[{node->left,true}] + mp[{node->right,true}] mp[{node, true}] = max(node->…

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called root. Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place…

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…

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. 示例 1: 输入: [1,2,3,1]输出: 4解释: 偷窃 1 号房屋 (金额 = 1) ,然后偷窃 3 号房屋 (金额 = 3).  偷窃到的最高金额 = 1 + 3 = 4 .示例 2: 输入: [2,…

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…

小偷又发现一个新的可行窃的地点. 这个地区只有一个入口,称为“根”. 除了根部之外,每栋房子有且只有一个父房子. 一番侦察之后,聪明的小偷意识到“这个地方的所有房屋形成了一棵二叉树”. 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警.在不触动警报的情况下,计算小偷一晚能盗取的最高金额.示例 1:     3    / \   2   3    \   \      3   1能盗取的最高金额 = 3 + 3 + 1 = 7.示例 2:     3    / \   4   5  / \…