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 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: [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: [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.
本题是LeetCode House Robber的扩展版，当首尾相连时，怎样偷到最多的钱。 因为首尾相连，所以如果偷了第一家，就不能偷最后一家；或者如果偷了最后一家，就不能偷第一家。所以我们可以把数组分成两个部分，一部分是去掉最后一家，求一个最大值；另一部分是去掉第一家，求一个最大值。最优结果就是这两个最大值的最大值。 代码如下：
class Solution {
public:
int rob(vector<int>& nums)
{
if (nums.size() == 0)
return 0;
if (nums.size() == 1)
return nums[0];
int ans1 = 0, ans2 = 0;
vector<vector<int> > dp1(nums.size() + 1, vector<int>(2, 0)), dp2(nums.size() + 1, vector<int>(2, 0));
for (int i = 1; i < nums.size(); ++i) { // 不偷最后一家
dp1[i][0] = max(dp1[i – 1][0], dp1[i – 1][1]);
dp1[i][1] = dp1[i – 1][0] + nums[i – 1];
ans1 = max(ans1, max(dp1[i][0], dp1[i][1]));
}
for (int i = 2; i <= nums.size(); ++i) { // 不偷第一家
dp2[i][0] = max(dp2[i – 1][0], dp2[i – 1][1]);
dp2[i][1] = dp2[i – 1][0] + nums[i – 1];
ans2 = max(ans2, max(dp2[i][0], dp2[i][1]));
}
return max(ans1, ans2);
}
};本代码提交AC，用时3MS。
更漂亮的代码是：
class Solution {
private:
int subrob(const vector<int>& nums, int left, int right)
{
int robsum = 0, norobsum = 0;
for (int i = left; i <= right; ++i) {
int nrs = max(robsum, norobsum);
int rs = norobsum + nums[i];
norobsum = nrs;
robsum = rs;
}
return max(robsum, norobsum);
}
public:
int rob(vector<int>& nums)
{
int n = nums.size();
if (n == 0)
return 0;
if (n == 1)
return nums[0];
return max(subrob(nums, 0, n – 2), subrob(nums, 1, n – 1));
}
};本代码提交AC，用时3MS。
二刷。常规解法：
class Solution {
private:
int rob_max(vector<int>& nums) {
int n = nums.size();
vector<int> dp(n, 0);
dp[0] = nums[0];
dp[1] = max(nums[0], nums[1]);
int ans = max(dp[0], dp[1]);
for (int i = 2; i < n; ++i) {
dp[i] = max(nums[i] + dp[i - 2], dp[i - 1]);
ans = max(ans, dp[i]);
}
return ans;
}
public:
int rob(vector<int>& nums) {
int n = nums.size();
if (n == 0)return 0;
if (n == 1)return nums[0];
if (n == 2)return max(nums[0], nums[1]);
vector<int> money1(nums.begin(), nums.end() - 1), money2(nums.begin() + 1, nums.end());
return max(rob_max(money1), rob_max(money2));
}
};本代码提交AC，用时4MS。