198 House Robber

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.

1 2 3 1

1 2 4 3

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.

后面的状态取决于前面的状态

money[i] = Math.max(money[i-2] + money[i], money[i-1])

time: O(n)

space: O(n)

If you want to save space, you could use two variables, one is prevYes, the other is prevNo, indicating whether you have picked the previous one or not.

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        if (nums.length == 1) {
            return nums[0];
        }
        int[] rob = new int[nums.length + 1];
        rob[0] = 0;
        rob[1] = nums[0];
        for (int i = 2 ; i <= nums.length; i++) {
            rob[i] = Math.max(rob[i - 2] + nums[i - 1], rob[i - 1]);
        }
        return rob[nums.length];
    }
}