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  • Java [Leetcode 337]House Robber III

    题目描述:

    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.

    解题思路:

    像House Robber I一样,使用动态规划法,对于每个节点,使用两个变量,res[0], res[1],分别表示不选择当前节点子树的数值和,选择当前节点子树的数值和,动态规划的思想,然后递归。

    代码如下:

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public int rob(TreeNode root) {
        	int[] res = robSub(root);
        	return Math.max(res[0], res[1]);
        }
    
        public int[] robSub(TreeNode root){
        	if(root == null)
        		return new int[2];
    
        	int[] left = robSub(root.left);
        	int[] right = robSub(root.right);
    
        	int[] res = new int[2];
        	res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]); // do not choose current node
        	res[1] = root.val + left[0] + right[0]; // choose current node
    
        	return res;
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zihaowang/p/5317718.html
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