Question

题目大意: 相当于在一列数组中取出一个或多个不相邻数，使其和最大(考虑收尾是相连接的)

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

Example 1:

Input: [2,3,2] Output: 3 Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.

Example 2:

Input: [1,2,3,1] Output: 4 Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.

Difficulty:Medium

Category:

Analyze

Solution

Solution 1: 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()));
 }

 private:
  int rob(vector<int>& nums, int left, int right) {
    int robEven = 0, robOdd = 0, n = nums.size();
    for (int i = left; i < right; ++i) {
      if (i % 2 == 0) {
        robEven = max(robEven + nums[i], robOdd);
      } else {
        robOdd = max(robEven, robOdd + nums[i]);
      }
    }

    return max(robEven, robOdd);
  }
};

Solution 2

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 rob = 0, notRob = 0;
    for (int i = left; i < right; ++i) {
      int preRob = rob, preNotRob = notRob;
      rob = preNotRob + nums[i];
      notRob = max(preRob, preNotRob);
    }
    return max(rob, notRob);
  }
};