House Robber II
最后更新于：2022-04-01 22:57:04
**一. 题目描述**
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.
**二. 题目分析**
该House Robber的升级版，只是增加了一点条件，即第一个element 和最后一个element不能同时出现，则需分为两种情况：
第一种情况: 不包括最后一个element；
第二种情况: 不包括第一个element；
两者的最大值即为全局最大值。
**三. 示例代码**
~~~
class Solution {
public:
int rob(vector& nums) {
if (nums.size() == 0) return 0;
if (nums.size() == 1) return nums[0];
return max(rob(nums, 0, nums.size()-1), rob(nums, 1, nums.size()));
}
int rob(vector &nums, int start, int end) {
int a = 0, b = 0;
for (int i = start; i < end; i++) {
int m = max(b, a+nums[i]);
a = b;
b = m;
}
return b;
}
};
~~~
**四. 小结**
无
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