House Robber && House RobberⅡ

House Robber && House RobberⅡ

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.

题目的意思：一个盗贼从一连串的房子里（可以看成数组）偷东西，不能偷相邻的房子（相邻的房子被偷会触发警报），最大能偷多少。转换：从给定的一个数组中，取数相加，相邻的不能取，能取的最大和是多少

思路，定义一个数组dp,dp[i]表示从1到i个数的能偷的最大值，对于i+1,我们知道，前i+1个房子（从1开始）最大值为前i个房子dp值和偷第i+1个房子加上dp[i-1]（需要绕开i个房子）两者的最大值，dp[i+1]=Max（dp[i],dp[i-1+nums[i]]）,dp[0]=0,dp[1]=nums[1],

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.empty())
            return 0;
        int len=nums.size();
        if(len==1)
            return nums[0];

        vector<int> dp(len+1,0);

        dp[1]=nums[0];
        for(int i=2;i<=len;i++)
        {
            dp[i]=Max(dp[i-1],nums[i-1]+dp[i-2]);
        }

        return dp[len];

    }

    int Max(int a,int b)
    {
        return a>b?a:b;
    }
};

House RobbereⅡ

现在房子的主人聪明了，把所有房子连成一个圈，即第一个房子和最后一个房子也是相邻的，问盗贼能偷的最大值是多少

思路:既然最后一个和第一个房子不能同时偷，那么先去掉最后一个房子，求 1 到 len-1的最大值，然后去掉第一个房子，求2到len个房子的最大值，去两者中 的 最大值

代码如下

class Solution {
public:
    int rob(vector<int>& nums) {

        if(nums.empty())
            return 0;
        int len=nums.size();
        if(len==1)
            return nums[0];

        return Max(rob(nums,0,len-1),rob(nums,1,len));
    }

    int rob(vector<int> &nums,int start,int end)
    {
        vector<int> dp(end+1,0);

        dp[start+1]=nums[start];
        dp[start]=0;

        for(int i=start+2;i<=end;i++)
        {
            dp[i]=Max(dp[i-1],dp[i-2]+nums[i-1]);
        }

        return dp[end];
    }

    int Max(int a,int b)
    {
        return a>b?a:b;
    }
};