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…

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. …

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…

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  / \…

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->…

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…

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…

337. 打家劫舍 III 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷意识到"这个地方的所有房屋的排列类似于一棵二叉树". 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. 计算在不触动警报的情况下,小偷一晚能够盗取的最高金额. 示例 1: 输入: [3,2,3,nu…