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 houses have security systems connected and it will automa
题目链接：https://leetcode.com/problems/house-robber/ 答案自然是动态规划(Dynamic programming，简称DP)。 代码 func rob1(nums []int) int { lth := len(nums) if lth == 0 { return 0 } dp := make([]int, lth+1) dp[1] = nums[0] for i := 2; i < lth+1; i++ {
198. House Robber Medium 10797236Add to ListShare 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
题目：题目链接 思路：看他的related topics。然后动态规划做。 这一看就是一维的，为什么因为二维的我不会，你也就看不到这篇题解了。嘿嘿 咱们令dp[i]表示前i个房子能偷的最大值。那么此时有两种情况：要么选第i个房子，那i-1就不能选，只能加上dp[i-2]。要么不选第i个房子，那么就可以是d
小偷问题-打家劫舍 这个问题同学面试的时候问到过，作为一道动态规划问题，就整理一下吧。 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
## 198. House RobberYou 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 systems connected a
这个题目是一个动态规划问题，每一步是上一步的最大值进行判断。 class Solution { public: int rob(vector<int>& nums) { if(nums.empty()) { return 0; } if(nums.size()==1) { return nums[0];
213. House Robber II 213. House Robber II题目思路复杂度代码 213. House Robber II 这道题是对198的简单扩展 Leetcode打家劫舍系列的对比可以参考Leetcode198. House Robber 题目 题目链接 思路 由于是环状,且相邻房屋不能同时被劫,只要保证第一个房屋和最后一个房
https://leetcode-cn.com/problems/house-robber/ 题目求数组非连续元素的最大和，状态转移方程为： dp[n] = max(dp[n-1],dp[n-2] + num) 所以本题只更新两个变量 ， dp[n-2] , dp[n-1] 为Fst Sec
并不是简单的奇偶位之和。 class Solution { public: int rob(vector<int>& nums) { if (nums.empty()) return 0; int size = nums.size(); if (size == 1) return nums[0]; int first = nums[0], second = max(nums[0], nums[1]);
题意描述 给定一个非负整数数组，找出其中累加和的最大值，并且相邻元素不能进行累加。 测试用例 Input: [1,2,3,1] Output: 4 Explanation: Total amount you can rob = 1 + 3 = 4. Input: [2,7,9,3,1] Output: 12 Explanation: Total amount you can rob = 2 + 9 + 1 = 12. 解题
leetcode 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 con
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 automatica
198. 打家劫舍 198. House Robber 每日一算法2019/5/8Day 5LeetCode198. 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
// DP class Solution { public: int rob(vector<int> &num) { if (num.size() <= 1) return num.empty() ? 0 : num[0]; vector<int> dp = {num[0], max(num[0], num[1])}; for (int i = 2; i < num.size(); ++i) {
