LeetCode周赛#105 Q2 Maximum Sum Circular Subarray (最大连续子列和变形题)

题目来源：https://leetcode.com/contest/weekly-contest-105/problems/maximum-sum-circular-subarray/

问题描述

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 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

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题意

求循环数组的最大连续子列和。与普通的最大连续子列和不同，这里的数组A是可以循环的，即从A[i]开始计算连续子列和，相当于计算数列B=A[i:end].append(A[0:i-1])的连续子列和。

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思路

传统的最大连续子列和用动态规划求解，dp[i]表示以i为结尾的最大连续子列和，复杂度为O(n)。如果直接将问题转化为n个不同数列的最大连续子列和求最大，则复杂度为O(n^2)，会超时。

解决方法比较有技巧性。循环数组的最大连续子列和=max(非循环数组的最大连续子列和，非循环数组的总和-非循环数组的最小连续子列和[但不能等于非循环数组的总和本身])，注意约束条件“非循环数组的最小连续子列和 不等于 非循环数组的总和本身”是为了避免出现子列长度为0的情况。这样原问题转化为两个复杂度为O(n)的子问题，总复杂度也是O(n).

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代码

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;
    }
};