标签:实现 算法 hashtable each roo htable into ken min
The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.
Determine the maximum amount of money the thief can rob tonight without alerting the police.
Example 1:
3 / 2 3 \ \ 3 1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3 / 4 5 / \ \ 1 3 1
Maximum amount of money the thief can rob = 4 + 5 = 9.
这道题本质上是与之前的两道题一样的。 但是区别在于这道题是Btree。 首先最好想的是NoRob与Rob两个method,分别计算。具体code如下
public int Rob(TreeNode root) { return Math.Max(Robb(root), NoRob(root)); } public int Robb(TreeNode root) { if(root == null) return 0; var left = NoRob(root.left); var right = NoRob(root.right); return left+right+root.val; } public int NoRob(TreeNode root) { if(root == null) return 0; return Math.Max(Robb(root.left), NoRob(root.left))+Math.Max(Robb(root.right), NoRob(root.right)); }
这个submit的时候超时了。问题在于里面Robb与NoRob两个method实际上repeat了很多次。需要改进算法,一种解决方法是输出是array。第一个数存的是恰好rob这个Node的,第二数存的是没有Rob这个node的最大值。最后只要比较root输出的这连个值,取最大值就好了。
具体代码为
public int Rob(TreeNode root) { var res = Robb(root); return Math.Max(res[0],res[1]); } public int[] Robb(TreeNode root) { var res = new int[2]; res[0] = 0; res[1] = 0; if(root == null) return res; var left = Robb(root.left); var right = Robb(root.right); res[0] = root.val+left[1]+right[1]; res[1] = Math.Max(left[0],left[1]) + Math.Max(right[0],right[1]); return res; }
此时想想能不能用第一种方法实现呢。理论上是可行的,即输入的时候加一个Hashtable,用来存每一个Node的极值。
标签:实现 算法 hashtable each roo htable into ken min
原文地址:http://www.cnblogs.com/renyualbert/p/5987055.html