House Robber II
題意：
與第一題相同，只是改成cycle
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.
解題思路：
其實與1相同，但是得分兩個情況來作：
- 搶第一間的話，得把最後一間去掉，去算max profit
- 搶最後一間的話，把第一間去掉，去算 max profit
最後再比較兩者哪個較大即可，注意邊際條件的處理，其程式碼如下：
public class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
int len = nums.length;
int robFirstHouse = rob(nums, 0, len - 1);
int robLastHouse = rob(nums, 1, len);
return Math.max(robFirstHouse, robLastHouse);
}
public int rob(int[] nums, int left, int right) {
if(left >= right) return 0;
int odd = 0;
int even = 0;
for (int i = left; i < right; i++) {
if (i % 2 == 1) {
odd = Math.max(odd + nums[i], even);
} else {
even = Math.max(odd, even + nums[i]);
}
}
return Math.max(even ,odd);
}
}