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Dynamic Programming - 213. House Robber II

Dynamic Programming - 213. House Robber II

213. House Robber II
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.

思路:

这一题和198题多了一个条件就是房子是排列成一个圆形,也就是说,在前一题的基础上,只能从0偷到len(nums)-2或者从1偷到len(num)-1。也是用动态规划求解。

代码:

func rob(nums []int) int {
    if nums == nil || len(nums) == 0{return 0}  
    leng := len(nums)
    if leng == 1 {return nums[0]}
    return max(helper(nums, 0, leng-2), helper(nums, 1, leng-1))  
}

func helper(input []int, start int, end int) int {
    nums := input[start:end+1]
    
    if nums == nil || len(nums) == 0{ return 0 }
    leng := len(nums)
    if leng == 1 {return nums[0]}
    
    dp := []int{nums[0], max(nums[0], nums[1])}
    index := 0
    for i := 2; i < leng; i++ {
        index = i % 2   //index^1: 0->1, 1->0.
        dp[index] = max(dp[index] + nums[i], dp[index^1])
    }
    
    return max(dp[0], dp[1])
} 

func max(i, j int) int {
    if i > j {
        return i 
    }
    return j
}

标题:Dynamic Programming - 213. House Robber II
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/08/14/1565714623038.html

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