198. 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 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.

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

题意

不能取数组中相邻的数,求能取到的最大值

分析

dp[i]表示从0到i最多能取多少, 对每个位置上的数,有取和不取两种选择,
如果取,说明前一个数不能取,这种情况下的值为dp[i−2]+nums[i],
如果不取, 前一个数可取可不取,直接使用dp[i−1]
状态转移方程:
dp[i]=max(dp[i−1],dp[i−2]+nums[i]);

代码

class Solution {public:    int rob(vector<int>& nums) {        int n = nums.size();        if (n <= 0) return 0;        if (n == 1) return nums[0];        vector<int> dp(n, 0);        dp[0] = nums[0];        dp[1] = nums[1] > nums[0]?nums[1]:nums[0];        for (int i = 2; i < nums.size(); i++) {            dp[i] = max(dp[i-1], dp[i-2]+nums[i]);        }        return dp[n-1];    }};