198.House Robber
198.House Robber
难度:Easy
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.
Example 1:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
Example 2:
Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.
要至少间隔1个，其实也可以看出，至多间隔两个。
简单暴力的方法，直接递归，四个一循环。
class Solution {
int rob(vector<int>& nums, int start) {
if(nums.empty()) return 0;
switch(nums.size() - start)
{
case 1: return nums[start];
case 2: return max(nums[start],nums[start+1]);
case 3: return max(nums[start]+nums[start+2],nums[start+1]);
case 4: return max(max(nums[start]+nums[start+2], nums[start+1]+nums[start+3]), nums[start]+nums[start+3]);
default: break;
}
return max(nums[start]+rob(nums,start+2), nums[start+1]+rob(nums, start+3));
}
public:
int rob(vector<int>& nums) {
return rob(nums,0);
}
};
不过会显示超时。
另一种方法，采用动态规划的方法，第n项的和可以看做是前n-2项加上第n项，与前n-1项和的较大值。
class Solution {
public:
int rob(vector<int>& nums) {
if(nums.empty()) return 0;
int prenum=0;
int curnum=0;
for(auto i:nums)
{
int tmp= prenum+i;
prenum=curnum;
curnum=max(tmp,curnum);
}
return max(curnum,prenum);
}
};
执行用时 :4 ms, 在所有 C++ 提交中击败了80.13%的用户 内存消耗 :8.4 MB, 在所有 C++ 提交中击败了93.46%的用户
Last modified 11mo ago