Leetcode# 198 House Robber

Implement the following operations of a stack using queues.

push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
empty() -- Return whether the stack is empty.

Notes:

You must use only standard operations of a queue -- which means only

push to back

,

peek/pop
from front

,

size

, and

is
empty

operations are valid.
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

Update (2015-06-11):

The class name of the Java function had been updated to MyStack instead of Stack.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and all test cases.

Difficulty:Easy

基本的思想是：nums[i] = max(nums[i-1], nums[i-2] + nums[i]);

需要注意的就是前四个数要提前确定下。

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]);
if(len>=4)
nums[3] = max(max(nums[0]+nums[2],nums[1]+nums[3]),nums[0]+nums[3]);
if(len>=3)
nums[2] = max(nums[0]+nums[2],nums[1]);

for(int i = 4;i<len;i++)
nums[i] = max(nums[i]+nums[i-2],nums[i-1]);

return max(nums[len-1],nums[len-2]);
}
};