LeetCode 213. House Robber II (Java版; Meidum)

梦里梦外; 2023-07-18 09:47 1阅读 0赞

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## LeetCode 213. House Robber II (Java版; Meidum) ##

#### 题目描述 ####

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.

#### 第一次做; 核心: 1)分两段使用动态规划, 一段是\[0,n-2\], 一段是\[1,n-1\] 2) LC198中三种DP含义 ####

class Solution { 
        public int rob(int[] nums) { 
            if(nums==null || nums.length==0){ 
                return 0;
            }
            if(nums.length==1){ 
                return nums[0];
            }
            int n = nums.length;
            //[0,n-2]
            int[][] dp = new int[n][2];
            dp[0][1] = nums[0];
            for(int i=1; i<n-1; i++){ 
                dp[i][0] = Math.max(dp[i-1][0], dp[i-1][1]);
                dp[i][1] = dp[i-1][0] + nums[i];
            }
            int max = Math.max(dp[n-2][0], dp[n-2][1]);
    
            //[1,n-1]
            int[][] dp2=  new int[n][2];
            dp2[1][1] = nums[1];
            for(int i=2; i<n; i++){ 
                dp2[i][0] = Math.max(dp2[i-1][0], dp2[i-1][1]);
                dp2[i][1] = dp2[i-1][0] + nums[i];
            }
            return Math.max(max, dp2[n-1][1]);
        }
    }

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