House Robber II

ID:213

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 a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Input: nums = [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.

Idea

Dynamic Programming + case consider

Since the result of the last house can either use index 0 house or not use index 0 house. We can consider two cases: 1. Start from 0 to num.length-2 index, which exclude the last index; 2. Start from 1 to last index, which exclude the first index

Keep track of two max values which are appliable to either odd index or even index. For example, dp[0] = nums[0] and dp[1] = nums[1]. Starting from index 2, current max value = Math.max(dp[0]+nums[2], dp[1]); Then dp[0]=dp[1] and dp[1] = current max.

Therefore, no adjacent index will use the max of adjacent index

Code

public int rob(int[] nums) {
    if(nums.length == 0) return 0;
    if(nums.length == 1) return nums[0];
    if(nums.length == 2) return Math.max(nums[0],nums[1]);        
    return Math.max(robHouses(nums, 0, nums.length-1),robHouses(nums, 1, nums.length));
}
    
    public int robHouses(int[] nums,int s, int e){
    int[] d = {nums[s], Math.max(nums[s], nums[s+1])};
    int cur = d[1];
    for(int i=s+2; i < e; i++){
        cur=Math.max(d[0]+nums[i], d[1]);
        d[0]=d[1];
        d[1]=cur;
    }
    return cur;
}