198. House Robber I and II

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

解题思路：

1. 计算当前index的sum值

代码如下：
class Solution {
public int rob(int[] nums) {

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

}

https://leetcode.com/problems/house-robber-iii/description/
解题思路：

1. 计算当前index的sum值，然后比较array最后index和index-1的sum值就好了
   class Solution {
   public int rob(int[] nums) {

    int len = nums.length;
    if(len <= 0) return 0;
    else if(len == 1) return nums[0];
    else if(len == 2) return Math.max(nums[0],nums[1]);
    int[] forward = new int[len];
    int[] back = new int[len];
    forward[0] = nums[0]; forward[1] = Math.max(nums[0],nums[1]);
    back[1] = nums[1]; back[2] = Math.max(nums[1],nums[2]);
    for(int i = 2; i < len - 1; i++){
        forward[i] = Math.max(forward[i-1], forward[i-2] + nums[i]);
    }
    for(int i = 3; i < len; i++){
        back[i] = Math.max(back[i-1], back[i-2] + nums[i]);
    }
    return Math.max(forward[len - 2], back[len-1]);
   

   }
   }