LeetCode学习篇十四——House Robber

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本文介绍了一种解决特定抢劫问题的方法，即如何在不触动报警系统的情况下从一系列房屋中窃取最大金额。通过动态规划算法，实现了高效求解。

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.

难度：easy 通过率：26.8%

理解了这道题的题意，状态转移方程就很容易想出来了，利用count数组记录偷取前i家的财产总额，如果偷取第i家，则加上count[i-2]，和偷取前i-1家的财产总额比较，取较大值。这里有个需要注意的地方，一开始忘记判断特殊情况当数组大小为0或1时，所以一直导致RTE。
代码实现如下，时间复杂度：O(n)

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size() == 0) return 0;
        if(nums.size() == 1) return nums[0];
        vector<int> count(nums.size(),0);
        count[0] = nums[0];
        count[1] = max(nums[0],nums[1]);
        for(int i = 2; i < nums.size(); i++) {
            count[i] = max(count[i-2]+nums[i], count[i-1]);
        }
        return count[nums.size()-1];
    }
};