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
动态计划
状态转移方程：f[i] =max(f[i⑴],f[i⑵]+c[i])
f[i]表示进入第i+1个房间时所得到的最大财富。
为了节省空间，只是用3个变量prepre,pre,cur便可。
实现代码1
//Runtime:2ms
class Solution {
public:
int rob(vector<int>& nums) {
int len = nums.size();
if (len == 0) return 0;
if (len == 1) return nums[0];
if (len == 2) return max(nums[0],nums[1]);
int prepre = nums[0];
int pre = max(nums[0],nums[1]);
int cur;
for (int i = 2; i < len; i++)
{
cur = max(pre,prepre + nums[i]);
prepre = pre;
pre = cur;
}
return cur;
}
};实现代码2
# Runtime:76ms
class Solution:
# @param {integer[]} nums
# @return {integer}
def rob(self,nums):
size = len(nums)
if size == 0:
return 0
elif size == 1:
return nums[0]
elif size == 2:
return max(nums[0],nums[1])
prepre = nums[0]
pre = max(nums[0],nums[1])
for i in range(2,size):
cur = max(pre,prepre + nums[i])
prepre = pre
pre = cur
return cur