[LeetCode] 918. Maximum Sum Circular Subarray 環形子數組的最大和

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:

1. -30000 <= A[i] <= 30000
2. 1 <= A.length <= 30000

這道題讓求環形子數組的最大和，對於環形數組，我們應該並不陌生，之前也做過類似的題目 [Circular Array Loop](http://www.cnblogs.com/grandyang/p/7658128.html)，就是說遍歷到末尾之后又能回到開頭繼續遍歷。假如沒有環形數組這一個條件，其實就跟之前那道 [Maximum Subarray](http://www.cnblogs.com/grandyang/p/4377150.html) 一樣，解法比較直接易懂。這里加上了環形數組的條件，難度就增加了一些，需要用到一些 trick。既然是子數組，則意味着必須是相連的數字，而由於環形數組的存在，說明可以首尾相連，這樣的話，最長子數組的范圍可以有兩種情況，一種是正常的，數組中的某一段子數組，另一種是分為兩段的，即首尾相連的，可以參見 [大神 lee215 的帖子](https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass) 中的示意圖。對於第一種情況，其實就是之前那道題 [Maximum Subarray](http://www.cnblogs.com/grandyang/p/4377150.html) 的做法，對於第二種情況，需要轉換一下思路，除去兩段的部分，中間剩的那段子數組其實是和最小的子數組，只要用之前的方法求出子數組的最小和，用數組總數字和一減，同樣可以得到最大和。兩種情況的最大和都要計算出來，取二者之間的較大值才是真正的和最大的子數組。但是這里有個 corner case 需要注意一下，假如數組中全是負數，那么和最小的子數組就是原數組本身，則求出的差值是0，而第一種情況求出的和最大的子數組也應該是負數，那么二者一比較，返回0就不對了，所以這種特殊情況需要單獨處理一下，參見代碼如下：

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

類似題目：

Maximum Subarray

Circular Array Loop

參考資料：

https://leetcode.com/problems/maximum-sum-circular-subarray/

https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass

[LeetCode All in One 題目講解匯總(持續更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)