leetcode 198. House Robber-优快云博客

本文通过动态规划解决了一道经典算法问题——打家劫舍。问题要求在不触动相邻房屋警报的情况下，计算出能从一系列房屋中窃取的最大金额。文中详细介绍了状态转移方程，并给出了一段实现该算法的Java代码。

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原文链接：http://bluereader.org/article/125400253

题目

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.

解

public class Solution {
    //解题思路：
//对于第i个房间我们的选择是偷和不偷， 如果决定是偷 则第i-1个房间必须不偷 那么 这一步的就是 dp[i] = nums(i-1) + dpNotTake[i -1] , 假设dp[i]表示打劫到第i间房屋时累计取得的金钱最大值.
//如果是不偷， 那么上一步就无所谓是不是已经偷过， dp[i] = dp[i -1 ]， 因此 dp[i] =max(dpNotTake[i-1 ] + nums(i-1), dp[i-1] );  其中dpNotTake[i-1]=dp[i-2]
//利用动态规划，状态转移方程：dp[i] = max(dp[i - 1], dp[i - 2] + num[i - 1])
//其中，dp[i]表示打劫到第i间房屋时累计取得的金钱最大值。
      public int rob(int[] nums) {
          int n = nums.length;
          if (n == 0) return 0;
          int[] dp = new int[n + 1];
          dp[0] = 0;
          dp[1] = nums[0];
          for (int i = 2; i < n + 1; ++i){
               dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i - 1]);
          }
          return dp[n]; 
    }
}