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.
Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.
方法一：递归（超时）
代码如下：
public class Solution {
public int rob(int[] nums) {
int m1= robWay(nums,0);
int m2= robWay(nums,1);
int max=m1>m2?m1:m2;
return max;
}
public int robWay(int[] nums,int begin){
int len=nums.length;
if(begin==len-1) return nums[begin];
if(begin>len-1) return 0;
int m1=nums[begin]+robWay(nums,begin+2);
int m2=nums[begin]+robWay(nums,begin+3);
int max=m1>m2?m1:m2;
return max;
}
}
运行结果：
方法2：动态规划
针对第n个房间，要么偷，要么不偷。
偷：take=noTake+nums[i];
不偷:noTake=maxProfile;
maxProfile=Math.max(take,noTake);
代码如下：
public class Solution {
public int rob(int[] nums) {
int take=0;
int noTake=0;
int maxProfit=0;
for(int i=0;i<nums.length;i++){
take=nums[i]+noTake;
noTake=maxProfit;
maxProfit=Math.max(take,noTake);
}
return maxProfit;
}
}
运行结果:
