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.
最開始用回溯寫,太暴力了超時了。
列舉各種選擇情況又不太現實。
思想是動態規劃,要掌握dp思想!!
class Solution {
public:
int rob(vector<int>& nums) {
if(nums.size()==0)
return 0;
if(nums.size()==1)
return nums[0];
if(nums.size()==2)
return max(nums[0],nums[1]);
vector<int > dp(nums.size());
dp[0]=nums[0];
dp[1]=max(nums[0],nums[1]);
for(int i=2;i<nums.size();++i)
{
dp[i]=max(nums[i]+dp[i-2],dp[i-1]);
}
return dp[dp.size()-1];
}
};