(算法分析Week14)House Robber[Easy]

198. House Robber[Easy]

题目来源

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 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.

Solution

给出一些非负数，代表抢劫每一间房子能得到的钱数，不能抢劫相邻的房子，求问能抢到的最大钱数是多少。
一个比较简单的动态规划问题。dp[i]表示到第i家的时候能抢到的钱的数目。由于不能连续抢劫，所以如果选择抢劫第i家，就不能抢劫第i-1家，所以dp[i] = nums[i] += dp[i-2];在dp[i]和dp[i-1]中选择最大的作为dp[i]的值，最后返回数组dp最末尾一个值即为抢劫nums.size()家能得到的最大钱数。

Complexity analysis

O(n)

Code

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

Result