213. House Robber II 首尾相同的偷窃问题

［抄题］：

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 houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),because they are adjacent houses.

Example 2:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).Total amount you can rob = 1 + 3 = 4.

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

不知道怎么处理首尾重复的问题：分情况讨论。从0-n-1, 1-n

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［一句话思路］：

exclude必须是用上一次的结果i e，否则会越加越大。所以要把上一次的结果用i e保存起来。

［输入量］：空：正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

分情况讨论也是一种办法。

［复杂度］：Time complexity: O(n) Space complexity: O(1)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

[是否头一次写此类driver funcion的代码] ：

[潜台词] ：

class Solution {public int rob(int[] nums) {//corner caseif (nums == null || nums.length == 0) return 0;if (nums.length == 1) return nums[0];//discuss in 2 waysreturn Math.max(rob(nums, 0, nums.length - 2), rob(nums, 1, nums.length - 1));}public int rob(int[] nums, int low, int high) {//define include, excludeint include = 0; int exclude = 0;//for loop, define i and e, and expandfor (int j = low; j <= high; j++) {int i = include; int e = exclude;include = e + nums[j];exclude = Math.max(i, e);}//return maxreturn Math.max(include, exclude);}
}

View Code

转载于:https://www.cnblogs.com/immiao0319/p/9515468.html

213. House Robber II 首尾相同的偷窃问题相关推荐

213. House Robber II
一.题目 1.审题 2.分析一个连续的数组,你能获取不连续的元素值代表的财富,数组首尾的两个元素算作连续,求能获得的最大财富值. 二.解答 1.思路: 在 198. House Robber 的基础 ...
LeetCode 213 House Robber II Python
题意: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 ...
leetcode 213. House Robber II | 213. 打家劫舍 II（Java）
题目 https://leetcode.com/problems/house-robber-ii/ 题解这道题是「198. 打家劫舍」的进阶,和第 198 题的不同之处是,这道题中的房屋是首尾相连的 ...
LeetCode 213. House Robber II（小偷游戏）
原题网址:https://leetcode.com/problems/house-robber-ii/ Note: This is an extension of House Robber. Afte ...
[leetcode] 213. House Robber II 解题报告
题目链接:https://leetcode.com/problems/house-robber-ii/ Note: This is an extension of House Robber. Afte ...
【Leetcode】213. 打家劫舍II（House Robber II）
Leetcode - 213 House Robber II (Medium) 题目描述:一个小偷沿着一条环形的街偷窃,给定数组表示每家屋子的金额,如果偷窃连续的两间屋子就会触发警报,求在不触发警报的 ...
LeetCode 198, 213 House Robber
198 House Robber DP 0~n-1 ans=dp[n-1] dp[i] = max(dp[i-2]+nums[i], dp[i-1]) i>=2 如果要输出偷了那些房子 ...
Leetcode题目：House Robber II
题目: Note: This is an extension of House Robber. After robbing those houses on that street, the thief ...
LeetCode OJ 之 House Robber II （抢劫犯 - 二）
题目: Note: This is an extension of House Robber. After robbing those houses on that street, the thief ...

213. House Robber II 首尾相同的偷窃问题

213. House Robber II 首尾相同的偷窃问题相关推荐

最新文章

热门文章