網址：https://leetcode.com/problems/house-robber/

參考：https://leetcode.com/problems/house-robber/discuss/156523/From-good-to-great.-How-to-approach-most-of-DP-problems.

容易得出轉移方程，然後要考慮各種極端情況。

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

這麽做代碼顯得稍微繁瑣，並且可以把dp數組簡化為3個不斷更新的變量！

class Solution {
    public int rob(int[] nums) {
        int pre1 = 0;
        int pre2 = 0;
        int temp = 0;
        
        for(int num : nums)
        {
            temp = Math.max(num + pre2, pre1);
            pre2 = pre1;
            pre1 = temp;
        }
        return temp;
    }
}

198. House Robber 強盜搶劫 Java