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.
分析
首先边界是dp[0]=nums[0],dp[1]=max(nums[0],nums[1]);对每一个house来说，只有抢与不抢两种情况，故动态转移方程dp[n]=max(dp[n-1],dp[n-2]+nums[n]).
class Solution {
public:
int rob(vector<int>& nums) {
if(nums.size()==0) return 0;
else if(nums.size()==1) return nums[0];
int dp[nums.size()];
dp[0]=nums[0]; dp[1]=max(nums[0],nums[1]);
for(int i=2;i<nums.size();i++)
dp[i]=max(dp[i-1],dp[i-2]+nums[i]);
return dp[nums.size()-1];
}
};
空间复杂度为O（1）的
class Solution {
public:
int rob(vector<int>& nums) {
int a=0,b=0;
for(int i=0;i<nums.size();i++){
if(i%2==0)
a=max(a+nums[i],b);
else
b=max(b+nums[i],a);
}
return max(a,b);
}
};