算法设计与应用基础： 第十周（1）

198. House Robber（动态规划的第一题）

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Total Accepted: 128389
Total Submissions: 335834
Difficulty: Easy
Contributor: LeetCode

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.

解题思路：类似Floyd算法，不过特殊之处在于有两种情况：对于每一座房子，选择抢或者不抢，对应于两个参数take，no_take，不抢的话，更新no_take为max（no_take,take)两者中的大者，抢的话，更新take=no_take+nums[i]（抢只能在上一座不抢的基础上）。最后返回take和no_take的大者。class Solution {
public:
int rob(vector<int>& nums) {
int take=0,no_take=0;
for(int i=0;i<nums.size();i++)
{
int temp=no_take;
no_take=max(take,no_take);
take=temp+nums[i];
}
return max(take,no_take);
}
};总结：动态规划两个步骤：对每一步骤状态的划分；在特定情况下对局部子状态的记录dp数组使得算法优化。