LeetCode House Robber II

原题链接在这里：https://leetcode.com/problems/house-robber-ii/

题目：

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.

题解：

是House Robber的进阶题。

比较从第一家到倒数第二家能偷最大值 和 把第二家到最后一家能偷最大值, 返回较大者。

corner case nums.length == 1, return nums[0].

Time Complexity: O(n). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int rob(int[] nums) {
 3         if(nums == null || nums.length == 0){
 4             return 0;
 5         }
 6         if(nums.length == 1){
 7             return nums[0];
 8         }
 9         return Math.max(robHelper(nums, 0, nums.length-2), robHelper(nums, 1, nums.length-1));
10     }
11     
12     private int robHelper(int [] nums, int l, int r){
13         int include = 0;
14         int exclude = 0;
15         for(int cur = l; cur<=r; cur++){
16             int i = include;
17             int e = exclude;
18             exclude = Math.max(i, e);
19             include = e+nums[cur];
20         }
21         return Math.max(exclude, include);
22     }
23 }

跟上House Robber III.

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/4824966.html