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.

递推关系我本来想用global和local做，但是想复杂了，不知怎么动态处理上一家有没有rob的情况。网上搜了dynamic transfer equation之后恍然大悟。。还是不够啊

递推关系为maxV[i] = max(maxV[i-2]+num[i], maxV[i-1]
maxV[i]表示到第i个房子位置，最大收益。

    public int rob(int[] nums) {
        if (nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        // maximum amount till the ith house
        int dp[] = new int[nums.length];
        //dp[i] = max(dp[i-2] + nums[i] , dp[i-1] )
        dp[0] = nums[0];
        dp[1] = nums[1] > nums[0] ? nums[1] : nums[0];
        for (int i = 2; i < nums.length; i++) {
            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
        }
        return dp[dp.length - 1];
    }