Leetcode:198. House Robber(week 9)

Description:

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：

题意为给定一个数组，数组的每个位置表示一个房子，数组的值为钱的数量，相邻的数不能同时获得，否则会惊动警察，其本质就是在一列数组中取出一个或多个不相邻数，使其和最大。 这是一道动态规划问题。 可以采用动态规划的方法实现。我采用的算法是遍历一遍，然后遍历的同时将不相邻的数加起来取最大值，时间复杂度为O(1),代码如下：

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) {
if (a+ nums[i] > b) {
a = a + nums[i];
} else {
a = b;
}
} else {
if (b+ nums[i] > a) {
b = b + nums[i];
} else {
b = a;
}
}
}
if (a > b) return a;
else return b;
}
};