LeetCode-198. House Robber [C++][Java]

LeetCode-198. House Robberhttps://leetcode.com/problems/house-robber/

题目描述

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 automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [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.

Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 400

解题思路

【C++】

动态规划：不抢劫当前这个房子dp[i-1], 或者抢劫当前这个房子nums[i-1] + dp[i-2]

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

压缩一下空间

class Solution {
public:
    int rob(vector& nums) {
        int pre2 = 0, pre1 = 0, len = nums.size();
        for(int i=0; i

【Java】

class Solution {
    public int rob(int[] nums) {
        if (nums.length == 0) {return 0;}
        int n = nums.length;
        int[] dp = new int[n + 1];
        dp[1] = nums[0];
        for (int i = 2; i <= n; ++i) {
            dp[i] = Math.max(dp[i-1], nums[i-1] + dp[i-2]);
        }
        return dp[n];
    }
}