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…

Note: This is an extension of House Robber. After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. …

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent…

注意事项: 这是 打家劫舍 的延伸.在上次盗窃完一条街道之后,窃贼又转到了一个新的地方,这样他就不会引起太多注意.这一次,这个地方的所有房屋都围成一圈.这意味着第一个房子是最后一个是紧挨着的.同时,这些房屋的安全系统与上次那条街道的安全系统保持一致.给出一份代表每个房屋存放钱数的非负整数列表,确定你可以在不触动警报的情况下盗取的最高金额. 详见:https://leetcode.com/problems/house-robber-ii/description/ Java实现: class Sol…

Note: This is an extension of House Robber. After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle.…

Note: This is an extension of House Robber. After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. …

原题链接 比子母题House Robber多了一个条件:偷了0以后,第n-1间房子不能偷. 转换思路为求偷盗[0,n-1)之间,以及[1,n)之间的最大值. 用两个DP,分别保存偷不偷第0间房的情况. Runtime: 0 ms, faster than 100.00% class Solution { public: int rob(vector<int> &nums) { ; int len = nums.size(); ) return res; ) ]; ) ] > nu…

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…

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…

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…