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.

遞推關(guān)系我本來想用global和local做，但是想復(fù)雜了，不知怎么動態(tài)處理上一家有沒有rob的情況。網(wǎng)上搜了dynamic transfer equation之后恍然大悟。。還是不夠啊

遞推關(guān)系為maxV[i] = max(maxV[i-2]+num[i], maxV[i-1]
maxV[i]表示到第i個房子位置，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];
    }