[LeetCode] 213. House Robber II 打家劫舍 II

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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

198. House Robber 的拓展，现在房子排成了一个圆圈，抢第一个屋子就不能抢最后一个屋子，抢最后一个屋子就不能抢第一个屋子。

解法：还是动态规划DP，分别算出这两个条件下的最大抢劫金额，然后取更大的就行。

Java:

public class Solution {

    public int rob(int[] nums) {
        return Math.max(rob(nums, 0), rob(nums, 1));
    }

    public int rob(int[] nums, int offset) {
        // 如果长度过小，则直接返回结果
        if(nums.length <= 1 + offset){
            return nums.length <= offset ? 0 : nums[0 + offset];
        }
        int a = nums[0 + offset];
        // 如果offset是1，则从下标为1的元素开始计算，所以要比较nums[1]和nums[2]
        int b = Math.max(nums[0 + offset], nums[1 + offset]);
        // 对于不抢劫最后一个房子的情况，i要小于nums.length - 1
        for(int i = 2 + offset; i < nums.length - 1 + offset; i++){
            int tmp = b;
            b = Math.max(a + nums[i], b);
            a = tmp;
        }
        return b;
    }
}

Python:

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

C++:　　

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        if (nums.size() == 1) {
            return nums[0];
        }

        return max(robRange(nums, 0, nums.size() - 1), // Include the first one of nums without the last one.
                   robRange(nums, 1, nums.size()));    // Include the last one of nums without the first one.
    }

    int robRange(vector<int>& nums, int start, int end) {
        int num_i = nums[start], num_i_1 = 0, num_i_2 = 0;
        for (int i = start + 1; i < end; ++i) {
            num_i_2 = num_i_1;
            num_i_1 = num_i;
            num_i = max(nums[i] + num_i_2, num_i_1);
        }
        return num_i;
    }
};

C++:

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() <= 1) return nums.empty() ? 0 : nums[0];
        return max(rob(nums, 0, nums.size() - 1), rob(nums, 1, nums.size()));
    }
    int rob(vector<int> &nums, int left, int right) {
        int a = 0, b = 0;
        for (int i = left; i < right; ++i) {
            int m = a, n = b;
            a = n + nums[i];
            b = max(m, n);
        }
        return max(a, b);
    }
};

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