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.
解决思路
动态规划.若偷，就是之前一个不偷的最大值加上当前房子的值，作为当前最大值；若不偷，之前一个既可以偷，也可以不偷，所以取之前偷或者不偷的最大值，作为当前的最大值。最终，取最后一个偷或者不偷的最大值，即为最终的结果。
代码
package leetcode;
public class HouseRobber {
public int rob(int[] nums) {
int rob = 0;
int notRob = 0;
for (int i : nums) {
int preMax = Math.max(rob, notRob);
rob = notRob + i;
notRob = preMax;
}
return Math.max(rob, notRob);
}
}
