问题描述：
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.
源码：
题目比较难懂，我先说一下意思，就是选一些值，让他们的和最大并且不能相邻。
用一个dp[i]表示前i个元素满足题意得最大值。递归方程如下：
dp[i] = max(result, dp[i-2]+nums[i])
时间复杂度O（n），空间O（n）。
时间空间都为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;
}
};
最后想了一下，其实可以不需要O（n），因为每次只需要用dp[i-2]和result的值，用两个变量存下来就行了。