leetcode 198. House Robber
leetcode 198. House Robber
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 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.
sulution
动态规划
对于第{i}家，你可以选择rob or not,如果rob那么第{i-1}家就不能rob，其值为dp[i-2]+nums[i],如果不rob 那么值就为dp[i-1]
其状态转移方程为
dp[i] = max{dp[i-1], dp[i-2] + nums[i]}
class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if(n == 0)
return 0;
if(n == 1)
return nums[0];
int dp[n+1];
dp[0] = nums[0];
dp[1] = nums[0] > nums[1] ? nums[0] : nums[1];
for(int i = 2; i < n; i++)
{
dp[i] = dp[i-1]>(dp[i-2] + nums[i]) ? dp[i-1] : dp[i-2] + nums[i];
}
return dp[n-1];
}
};
时间复杂度：O(N)
空间复杂度：O(N)
