題目來源：https://leetcode.com/contest/weekly-contest-105/problems/maximum-sum-circular-subarray/

問題描述

918. Maximum Sum Circular Subarray

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

Also, a subarray may only include each element of the fixed buffer A at most once.  (Formally, for a subarray 

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

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題意

求迴圈陣列的最大連續子列和。與普通的最大連續子列和不同，這裡的陣列A是可以迴圈的，即從A[i]開始計算連續子列和，相當於計算數列B=A[i:end].append(A[0:i-1])的連續子列和。

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思路

傳統的最大連續子列和用動態規劃求解，dp[i]表示以i為結尾的最大連續子列和，複雜度為O(n)。如果直接將問題轉化為n個不同數列的最大連續子列和求最大，則複雜度為O(n^2)，會超時。

解決方法比較有技巧性。迴圈陣列的最大連續子列和=max(非迴圈陣列的最大連續子列和，非迴圈陣列的總和-非迴圈陣列的最小連續子列和[但不能等於非迴圈陣列的總和本身])，注意約束條件“非迴圈陣列的最小連續子列和 不等於 非迴圈陣列的總和本身”是為了避免出現子列長度為0的情況。這樣原問題轉化為兩個複雜度為O(n)的子問題，總複雜度也是O(n).

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程式碼

class Solution {
public:
    int dp[30005];
    
    int maxSubArray(vector<int>& q)
    {
        int i, len = q.size(), vmax = -1000000000;
        memset(dp, 0, sizeof(dp));
        dp[0] = q[0];
        for (i=1; i<len; i++)
        {
            if (dp[i-1] >= 0)
            {
                dp[i] = q[i] + dp[i-1];
            }
            else
            {
                dp[i] = q[i];
            }
        }
        for (i=0; i<len; i++)
        {
            vmax = max(vmax, dp[i]);
        }
        return vmax;
    }
    int minSubArray(vector<int>& q)
    {
        int i, len = q.size(), vmin = 1000000000;
        memset(dp, 0, sizeof(dp));
        dp[0] = q[0];
        for (i=1; i<len; i++)
        {
            if (dp[i-1] <= 0)
            {
                dp[i] = q[i] + dp[i-1];
            }
            else
            {
                dp[i] = q[i];
            }
        }
        for (i=0; i<len; i++)
        {
            vmin = min(vmin, dp[i]);
        }
        return vmin;
    }
    int sumArray(vector<int>& q)
    {
        int i, len = q.size(), ans = 0;
        for (i=0; i<len; i++)
        {
            ans += q[i];
        }
        return ans;
    }
    int maxSubarraySumCircular(vector<int>& A) {
        int vmax = -1000000000, sum = sumArray(A), vmin = minSubArray(A);
        vmax = max(vmax, maxSubArray(A));
        if (sum != vmin)
        {
            vmax = max(vmax, sum - vmin);
        }
        return vmax;
    }
};