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  • 337. House Robber III

    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.

    题目含义:不能连续抢直接相连的两个节点。即例2中,抢了3就不能抢4,5。问最多能抢好多。

    思路:给一个二叉树,求它不直接相连的节点的val和最大为多少。

    如果抢了当前节点,那么它的左右孩子就肯定不能抢了。 
    如果没有抢当前节点,左右孩子抢不抢取决于左右孩子的孩子的val大小。

     1     private int[] dfs(TreeNode root)
     2     {
     3         int[] result = {0,0}; //result[0]表示抢当前节点 result[1]表示不抢当前节点
     4         if (root == null) return result;
     5         int[] leftResult = dfs(root.left);
     6         int[] rightResult = dfs(root.right);
     7         result[0] = leftResult[1] +rightResult[1] + root.val;//抢了当前节点,它的左右孩子就不可以抢了
     8         result[1] = Math.max(leftResult[0],leftResult[1]) + Math.max(rightResult[0],rightResult[1]);//不抢当前节点,左右孩子可抢可不抢
     9         return result;
    10     }
    11     
    12     public int rob(TreeNode root) {
    13         int[] result = dfs(root);
    14         return Math.max(result[0],result[1]);        
    15     }
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  • 原文地址:https://www.cnblogs.com/wzj4858/p/7712050.html
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