House Robber II LeetCode Java

本博客参考自：https://leetcode.com/discuss/36544/simple-ac-solution-in-java-in-o-n-with-explanation

Note: This is an extension of House Robber.

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

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.

这道题目是House Robber的延伸。House Rouber的意思是，一个偷，找到了条街道，然后准备偷钱，这个街道的报警系统比较傻瓜，只要连续两家不同时被偷，就不会报警，求小偷能偷到的最大钱数。

而House Robber II和之前的差别在于，街道编程了一个环，如果你抢了第一家，就不能抢最后一家。

两道题目思路都是使用动态规划。

首先看House Robber。

传统思路：使用数组int[]作为记录数组，int[i] 表示偷到第i家，并且偷第i家的情况，偷到钱最大数。那么有：

money[i]=Max(money[i-2], money[i-3])+nums[i];

但是这样需要额外的O（n）的存储空间，优化后，可使用O（1）空间完成，程序如下（未考虑边界情况）：

private int rob(int[] num) {
int include = 0, exclude = 0;
for (int j =0; j <nums.length; j++) {
int i = include, e = exclude;
include = e + num[j];
exclude = Math.max(e, i);
}
return Math.max(include, exclude);
}

接下来考虑House Robber II 的问题

考虑这道题目的实际情况，假设有n个住户，那么至少有n/2个用户不被抢。而且任何一个用户要么被抢，要么不被抢。如果第i个用户不被抢，那么可以看做环从第i个位置断开，如果第i个用户被抢，则第i+1个用户一定不被抢，可以看做环从第i+1个位置断开。当然，i是任意的。

所以，可以将House RobberII 转换为 House Robber的问题。

程序如下（这里i和i+1 分别为length-1 和 0）：

public int rob(int[] nums) {
if (nums.length == 1) return nums[0];
return Math.max(rob(nums, 0, nums.length - 2), rob(nums, 1, nums.length - 1));
}

private int rob(int[] num, int lo, int hi) {
int include = 0, exclude = 0;
for (int j = lo; j <= hi; j++) {
int i = include, e = exclude;
include = e + num[j];
exclude = Math.max(e, i);
}
return Math.max(include, exclude);
}