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.
動態(tài)計劃
狀態(tài)轉移方程：f[i] =max(f[i⑴], f[i⑵]+c[i])
f[i]表示進入第i+1個房間時所得到的最大財富。
為了節(jié)省空間，只是用3個變量prepre, pre, cur便可。
//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;
}
};# 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
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