算法分析与设计——LeetCode Problem.213 House Robber II

题目链接

问题描述

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.

解题思路

与Leetcode 198 题不同的是这题第一个房间与最后一个房间相连，即形成了一个环。那么假如选择第一个房间，那最后一个房间就不能进入；如果不选择第一个房间，就存在进入最后一个房间的可能性。然后将这两种情况得出的最大值进行比较，即得到最优解。

代码如下

class Solution {
public:
int rob(vector<int>& nums) {
if (nums.size() == 0) return 0;
if (nums.size() == 1) return nums[0];
if (nums.size() == 2) return max(nums[0], nums[1]);

int leng = nums.size();

vector<int> vec1(leng - 1, 0);
vector<int> vec2(leng, 0);

vec1[0] = nums[0];
vec1[1] = max(nums[0], nums[1]);

vec2[1] = nums[1];
vec2[2] = max(nums[1], nums[2]);

for (int i = 2; i < nums.size() - 1; i++) {
vec1[i] = max(vec1[i - 1], vec1[i - 2] + nums[i]);
}

for (int i = 3; i < nums.size(); i++) {
vec2[i] = max(vec2[i - 1], vec2[i - 2] + nums[i]);
}

return max(vec1[leng - 2], vec2[leng - 1]);
}
};