打家劫舍（House Robber）是LeetCode上比较典型的一个题目，涉及三道题，主要解题思想是动态规划，将三道题依次记录如下：
（一）打家劫舍
题目等级：198、House Robber（Easy）
题目描述：
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.Example 2:
Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.题意：你是一个专业的小偷，计划偷窃沿街的房屋。每间房内都藏有一定的现金，影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。给定一个代表每个房屋存放金额的非负整数数组，计算你在不触动警报装置的情况下，能够偷窃到的最高金额。
解题思路：
本题实际就是典型的动态规划问题，沿街的房屋依次决定每一家偷还是不偷为一个阶段，用dp[i]表示前i家获取的最高金额，第i阶段的决策就是两种：偷、不偷。则不难写出以下状态转移方程：
dp[i]=max(dp[i-1],dp[i-2]+nums[i])
class Solution {
public int rob(int[] nums) {
//典型的动态规划问题，dp[i]=max(dp[i-1],dp[i-2]+nums[i])
//每个阶段确定一家偷还是不偷，所以决策就是偷和不偷两种
if(nums==null || nums.length==0)
return 0;
int len=nums.length;
int[] res=new int[len+1];
res[0]=0;
res[1]=nums[0];
for(int i=2;i<=len;i++){
res[i]=Math.max(res[i-1],res[i-2]+nums[i-1]); //状态转移
}
return res[len];
}
}（二）打家劫舍 II
题目等级：213、House Robber II（Medium）
题目描述：
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
because they are adjacent houses.Example 2:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.题意：你是一个专业的小偷，计划偷窃沿街的房屋，每间房内都藏有一定的现金。这个地方所有的房屋都围成一圈，这意味着第一个房屋和最后一个房屋是紧挨着的。同时，相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。
解题思路：
本题和上题基本类似，区别在于：所有的房子围成一个圈，意味着首尾两家也属于相邻。
基本思路当然还是动态规划，可以划分成两种情况来做，以第一家是否被偷为依据分成两个动态规划问题，如果第一家偷，那么从第一家到第n-1家求最大值（因为意味着最后一家一定不能偷）；如果第一家不偷，那么从第2家到第n家求最大值。最后再比较两种情况的最大值即可。
class Solution {
public int rob(int[] nums) {
/*
思路：由于首尾也属于相邻，因此需要分别判断，以第一家是否打劫分成两个问题
第一家抢：最后一家一定不能抢，从第0个到len-2做动态规划
第一家不抢：从1到len-1做动态规划
然后比较找出最大值
*/
if(nums==null || nums.length==0)
return 0;
int len=nums.length;
if(len==1)
return nums[0];
int[] dp1=new int[len];
int[] dp2=new int[len+1];
//第一家抢
dp1[0]=0;
dp1[1]=nums[0];
for(int i=2;i<len;i++)
dp1[i]=Math.max(dp1[i-1],dp1[i-2]+nums[i-1]);
//第一家不抢
dp2[0]=0;
dp2[1]=0;
for(int i=2;i<=len;i++)
dp2[i]=Math.max(dp2[i-1],dp2[i-2]+nums[i-1]);
return Math.max(dp1[len-1],dp2[len]);
}
}（三）打家劫舍 III
题目等级：337、House Robber III（Medium）
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:
Input: [3,2,3,null,3,null,1]
3
/ \
2 3
\ \
3 1
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.Example 2:
Input: [3,4,5,1,3,null,1]
3
/ \
4 5
/ \ \
1 3 1
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.题意：在上次打劫完一条街道之后和一圈房屋后，小偷又发现了一个新的可行窃的地区。这个地区只有一个入口，我们称之为“根”。 除了“根”之外，每栋房子有且只有一个“父“房子与之相连。一番侦察之后，聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”。 如果两个直接相连的房子在同一天晚上被打劫，房屋将自动报警。
解题思路：
和上两题不同的是：本题是二叉树形状，树中直接相连（也就是父子节点）不能同时打劫。
实际上，有上一题的经验，我们不难找到规律：这里可以按根结点是否打劫分为两种情况，如果根结点被打劫，那么意味着它的子节点都不能打劫，需要直接打劫其孙子辈节点；如果根结点不打劫，那么可以考虑其左右孩子。同时，由于树的特性，左右子树都是子问题，因此实际上就是两种情况分别递归。
class Solution {
public int rob(TreeNode root) {
/*
按根偷不偷，分两种情况，分别递归
*/
if(root==null)
return 0;
int res1=0;
//根在结果中
res1+=root.val;
if(root.left!=null)
res1+=(rob(root.left.left)+rob(root.left.right));
if(root.right!=null)
res1+=(rob(root.right.left)+rob(root.right.right));
//根不在结果中
int res2=rob(root.left)+rob(root.right);
return Math.max(res1,res2);
}
}