問題描述：
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 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.
源碼：
題目比較難懂，我先說一下意思，就是選一些值，讓他們的和最大並且不能相鄰。
用一個dp[i]表示前i個元素滿足題意得最大值。遞歸方程如下：
dp[i] = max(result, dp[i-2]+nums[i])
時間複雜度O（n），空間O（n）。
時間空間都爲100%
class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if(n==0) return 0;
if(n==1) return nums[0];
vector<int> dp(n, 0);
dp[0] = nums[0];
dp[1] = max(nums[0], nums[1]);
int result = dp[1];
for(int i=2; i<n; i++){
dp[i] = max(result, dp[i-2]+nums[i]);
result = max(dp[i], result);
}
return result;
}
};
最後想了一下，其實可以不需要O（n），因爲每次只需要用dp[i-2]和result的值，用兩個變量存下來就行了。