[LeetCode] House Robber II 打家劫舍之二

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.

这道题是之前那道House Robber 打家劫舍的拓展,现在房子排成了一个圆圈,则如果抢了第一家,就不能抢最后一家,因为首尾相连了,所以第一家和最后一家只能抢其中的一家,或者都不抢,那我们这里变通一下,如果我们把第一家和最后一家分别去掉,各算一遍能抢的最大值,然后比较两个值取其中较大的一个即为所求。那我们只需参考之前的House Robber 打家劫舍中的解题方法,然后调用两边取较大值,代码如下:

解法一

// DP
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) {
        if (right - left <= 1) return nums[left];
        vector<int> dp(right, 0);
        dp[left] = nums[left];
        dp[left + 1] = max(nums[left], nums[left + 1]);
        for (int i = left + 2; i < right; ++i) {
            dp[i] = max(nums[i] + dp[i - 2], dp[i - 1]);
        }
        return dp.back();
    }
};

解法二

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);
    }
};
时间: 05-19

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