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, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will autom…

原题链接在这里:https://leetcode.com/problems/house-robber-iii/ 题目: 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 hous…

原题链接在这里:https://leetcode.com/problems/house-robber/ 题目: 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…

本题和House Robber差不多,分成两种情况来解决.第一家是不是偷了,如果偷了,那么最后一家肯定不能偷. class Solution(object): def rob(self, nums): """ :type nums: List[int] :rtype: int """ if not nums: return 0 n = len(nums) if n == 1: return nums[0] if n == 2: return ma…

213. House Robber II     Total Accepted: 24216 Total Submissions: 80632 Difficulty: Medium 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…

Description: 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…

题意:有一个整数序列,从中挑出一些数字,使得总和是最大,前提是,相邻的两个数字中只能挑其一.比如1 2 3 就只能挑2或者1和3. 思路:很直观的题,dp思想.降低规模,从小规模开始考虑.如果只有两个数字,那么结果很明显就能知道是其中之大者.假如已经知道了第 i 个之前的决策,那么第i+2个之前的决策也就知道了.前两个数字已经由人工得知,设为dp[0]和dp[1],那么dp[2]=max(dp[0]+nums[2], dp[1]).状态转移方程dp[i]=max(dp[i-1], dp[i-2]…

Here is my collection of solutions to leetcode problems. Related code can be found in this repo: https://github.com/zhuli19901106/leetcode LeetCode - Course Schedule LeetCode - Reverse Linked List LeetCode - Isomorphic Strings LeetCode - Count Primes…