Leetcode 213. House Robber II

https://leetcode.com/problems/house-robber-ii/

The majority part is still the same. Actually the difference is clear: if we rob first house, we cannot rob the last house.

Solution:

# T:O(n) S:O(1)
class Solution:
    # @param {integer[]} nums
    # @return {integer}
    def rob(self, nums):
        if len(nums) == 0:
            return 0
            
        if len(nums) == 1:
            return nums[0]
        
        return max(self.robRange(nums, 0, len(nums) - 1),\
                   self.robRange(nums, 1, len(nums)))
    
    def robRange(self, nums, start, end):
        num_i, num_i_1 = nums[start], 0
        for i in xrange(start + 1, end):
            num_i_1, num_i_2 = num_i, num_i_1
            num_i = max(nums[i] + num_i_2, num_i_1);
        
        return num_i

Run Time: 36 ms