You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a 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 nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: nums = [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
Example 2:
Input: nums = [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 3:
Input: nums = [0]
Output: 0
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 1000
class Solution {
public:
int rob(vector<int>& nums) {
int length = nums.size();
if (length == 1) {
return nums[0];
}
vector<int> result_1{0,0,0};
for (int i=0; i<length-1; ++i) {
int v2 = nums[i] + result_1[i-2+3];
int v3 = nums[i] + result_1[i-3+3];
result_1.emplace_back(std::max(v2, v3));
}
vector<int> result_2{0,0,0};
for (int i=1; i<length; ++i) {
int v2 = nums[i] + result_2[i-3+3];
int v3 = nums[i] + result_2[i-4+3];
result_2.emplace_back(std::max(v2, v3));
}
int ret = std::max({
result_1[length],
result_1[length+1],
result_2[length],
result_2[length+1],
});
return ret;
}
};
Runtime: 4 ms, faster than 48.77% of C++ online submissions for House Robber II.
Memory Usage: 8.4 MB, less than 5.19% of C++ online submissions for House Robber II.
一遍过,开心,但是性能一般,提升的话可以不用vector吧(用stack?),或者reserve vector(试了,没效果)
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