198-House Robber

类别：dynamic programming
难度：easy

题目描述

https://leetcode.com/problems/house-robber/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.

算法分析

（1）n == 0, return 0
（2）nums[0] = nums[2] = 原值
（3）nums[3] = max(nums[1], nums[0] + nums[2])
（4）nums[i] = max(nums[i] + nums[i - 2], nums[i] + nums[i - 3], nums[i - 1]);

代码实现

//  find maximum sum that do not choose the adjactive value at the same time //  leetcode 198#include <iostream>#include <vector>using namespace std;int rob(vector<int>& nums) {    int result = 0;    int n = nums.size();    if (n == 0) return 0;    nums[1] = max(nums[0], nums[1]);    nums[2] = max(nums[1], (nums[0] + nums[2]));    for (int i = 3; i < n; ++i) {        nums[i] = max(nums[i] + nums[i - 2], nums[i] + nums[i - 3]);        nums[i] = max(nums[i], nums[i - 1]);    }            return nums[n - 1];}int main() {    int n;    int num;    vector<int> data;    cin >> n;    for (int i = 0; i < n; ++i) {        cin >> num;        data.push_back(num);    }    cout << rob(data) << endl;    return 0;}