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.
思路：
動態規劃DP。本質相當於在一列數組中取出一個或多個不相鄰數，使其和最大。
State: dp[i]，表示到第i個房子時能夠搶到的最大金額。
Function: dp[i] = max(num[i] + dp[i - 2], dp[i - 1])
Initialize: dp[0] = num[0], dp[1] = max(num[0], num[1]) 或者 dp[0] = 0, dp[1] = 0
Return: dp[n]
c++代碼：
class Solution {
public:
int rob(vector<int>& nums) {
int dp = 0, dp_1 = 0, dp_2 = 0;//dp_1 表示dp[i-1], dp_2 表示dp[i-2]
int temp = 0;
for (int i = 0; i < nums.size(); i++) {
temp = dp_1;
dp = max(nums[i] + dp_2, dp_1);
dp_1 = dp;
dp_2 = temp;
}
return dp;
}
};