[LeetCode] 918. Maximum Sum Circular Subarray 环形子数组的最大和
Given a circular array C of integers represented by `A`, find the maximum possible sum of a non-empty subarray of C.
Here, a circular array means the end of the array connects to the beginning of the array. (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)
Also, a subarray may only include each element of the fixed buffer A at most once. (Formally, for a subarray C[i], C[i+1], ..., C[j], there does not exist i <= k1, k2 <= j with k1 % A.length = k2 % A.length.)
Example 1:
Input: [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3
Example 2:
Input: [5,-3,5]
Output: 10 Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10
Example 3:
Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4
Example 4:
Input: [3,-2,2,-3]
Output: 3 Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3
Example 5:
Input: [-2,-3,-1]
Output: -1 Explanation: Subarray [-1] has maximum sum -1
Note:
-30000 <= A[i] <= 300001 <= A.length <= 30000
这道题让求环形子数组的最大和，对于环形数组，我们应该并不陌生，之前也做过类似的题目 [Circular Array Loop](http://www.cnblogs.com/grandyang/p/7658128.html)，就是说遍历到末尾之后又能回到开头继续遍历。假如没有环形数组这一个条件，其实就跟之前那道 [Maximum Subarray](http://www.cnblogs.com/grandyang/p/4377150.html) 一样，解法比较直接易懂。这里加上了环形数组的条件，难度就增加了一些，需要用到一些 trick。既然是子数组，则意味着必须是相连的数字，而由于环形数组的存在，说明可以首尾相连，这样的话，最长子数组的范围可以有两种情况，一种是正常的，数组中的某一段子数组，另一种是分为两段的，即首尾相连的，可以参见 [大神 lee215 的帖子](https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass) 中的示意图。对于第一种情况，其实就是之前那道题 [Maximum Subarray](http://www.cnblogs.com/grandyang/p/4377150.html) 的做法，对于第二种情况，需要转换一下思路，除去两段的部分，中间剩的那段子数组其实是和最小的子数组，只要用之前的方法求出子数组的最小和，用数组总数字和一减，同样可以得到最大和。两种情况的最大和都要计算出来，取二者之间的较大值才是真正的和最大的子数组。但是这里有个 corner case 需要注意一下，假如数组中全是负数，那么和最小的子数组就是原数组本身，则求出的差值是0，而第一种情况求出的和最大的子数组也应该是负数，那么二者一比较，返回0就不对了，所以这种特殊情况需要单独处理一下，参见代码如下：
class Solution {
public:
int maxSubarraySumCircular(vector<int>& A) {
int sum = 0, mn = INT_MAX, mx = INT_MIN, curMax = 0, curMin = 0;
for (int num : A) {
curMin = min(curMin + num, num);
mn = min(mn, curMin);
curMax = max(curMax + num, num);
mx = max(mx, curMax);
sum += num;
}
return (sum - mn == 0) ? mx : max(mx, sum - mn);
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/918
类似题目：
参考资料：
https://leetcode.com/problems/maximum-sum-circular-subarray/
https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass
[LeetCode All in One 题目讲解汇总(持续更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)
