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.

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.

//dp: Time: O(n), Space: O(n)  
  public int rob(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        if (nums.length == 1) {//不要忘记对一个数的数组单独判断
            return nums[0];
        }
        
        int[] dp = new int[nums.length];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);
        
        for (int i = 2; i < nums.length; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        
        return dp[nums.length - 1];
    }
    //dp: Time:O(n), Space:O(1),a和b交替进行计算最大值
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        if (nums.length == 1) {//不要忘记对一个数的数组单独判断
            return nums[0];
        }
        
        int a = 0;//开始一定a和b归零，否则就没有把前两位的情况加到结果里
        int b = 0;
        
        for (int i = 0; i < nums.length; i++) {
            if (i % 2 == 0) {
                a = Math.max(a + nums[i], b);
            } else {
                b = Math.max(b + nums[i], a);
            }
        }
        
        return Math.max(a, b);
    }