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Dynamic Progamming - 198. House Robber

Dynamic Progamming - 198. House Robber

198. House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that 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: [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.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
  Total amount you can rob = 2 + 9 + 1 = 12.

思路:

题目意思是说,假如你是一个强盗,连着偷一排房子,不能连续偷,只能偷隔一家,问能偷的最多多少钱。
最后一步是偷或者不偷最后这一家所得到的钱最多是多少,而偷不偷最后一家取决于前面一家偷或者不偷,所以子问题就出来了,状态方程:dp[i] = max(dp[i-1] + nums[i], dp[i-2])这其实也是斐波那契数列,当然初始条件就是0和1,不能用状态转移方程求出来。

代码:

func rob(nums []int) int {
    if nums == nil || len(nums) == 0 { return 0 }
    if len(nums) == 1 { return nums[0] }  
    
    dp := []int{nums[0], max(nums[0], nums[1])}
    index := 0
    for i:=2; i < len(nums); i++ {
            index = i&1; // i&1 === i%2.  //index^1: 0->1, 1->0.
        	dp[index] = max(nums[i]+dp[index], 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 Progamming - 198. House Robber
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/08/13/1565709098978.html

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