198 House Robber 打家劫舍
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.
Example:
Example 1:
Input: [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: [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.
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，系统会自动报警。
给定一个代表每个房屋存放金额的非负整数数组，计算你在不触动警报装置的情况下，能够偷窃到的最高金额。
示例：
示例 1:
输入: [1,2,3,1]
输出: 4
解释: 偷窃 1 号房屋 (金额 = 1) ，然后偷窃 3 号房屋 (金额 = 3)。
偷窃到的最高金额 = 1 + 3 = 4 。
示例 2:
输入: [2,7,9,3,1]
输出: 12
解释: 偷窃 1 号房屋 (金额 = 2), 偷窃 3 号房屋 (金额 = 9)，接着偷窃 5 号房屋 (金额 = 1)。
偷窃到的最高金额 = 2 + 9 + 1 = 12 。
思路:
动态规划
状态转移公式为: f(n) = max[f(n - 1), f(n - 2) + an]
如果采用数组保存, 空间复杂度为O(n)(Java代码, 自顶向下/C++代码, 自底向上)
从状态转移公式可以看出, 实际上用到的只有 2个变量, 可以将空间进一步压缩
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int rob(vector& nums)
{
int len = nums.size();
if (len == 0) return 0;
if (len == 1) return nums[0];
vector memo(len);
memo[0] = nums[0];
memo[1] = nums[0] > nums[1] ? nums[0] : nums[1];
for (int i = 2; i < len; i++) memo[i] = max(memo[i - 1], memo[i - 2] + nums[i]);
return memo[len - 1];
}
};
Java:
class Solution {
private int[] memo;
public int rob(int[] nums) {
memo = new int[nums.length];
Arrays.fill(memo, -1);
return tryRob(nums, 0);
}
private int tryRob(int[] nums, int index) {
if (index >= nums.length) {
return 0;
}
if (memo[index] != -1) {
return memo[index];
}
int result = 0;
for (int i = index; i < nums.length; i++) {
result = Math.max(result, nums[i] + tryRob(nums, i + 2));
}
memo[index] = result;
return result;
}
}
Python:
class Solution:
def rob(self, nums: List[int]) -> int:
pre, result = 0, 0
for num in nums:
pre, result = result, max(num + pre, result)
return result