A non-empty zero-indexed array A consisting of N integers is given.

A triplet (X, Y, Z), such that 0 ≤ X < Y < Z < N, is called adouble slice.

Thesumof double slice (X, Y, Z) is the total of A[X + 1] + A[X + 2] + ... + A[Y − 1] + A[Y + 1] + A[Y + 2] + ... + A[Z − 1].

For example, array A such that:

    A[0] = 3
    A[1] = 2
    A[2] = 6
    A[3] = -1
    A[4] = 4
    A[5] = 5
    A[6] = -1
    A[7] = 2

contains the following example double slices:

• double slice (0, 3, 6), sum is 2 + 6 + 4 + 5 = 17,
• double slice (0, 3, 7), sum is 2 + 6 + 4 + 5 − 1 = 16,
• double slice (3, 4, 5), sum is 0.

The goal is to find the maximal sum of any double slice.

Write a function:

int solution(int A[], int N);

that, given a non-empty zero-indexed array A consisting of N integers, returns the maximal sum of any double slice.

For example, given:

    A[0] = 3
    A[1] = 2
    A[2] = 6
    A[3] = -1
    A[4] = 4
    A[5] = 5
    A[6] = -1
    A[7] = 2

the function should return 17, because no double slice of array A has a sum of greater than 17.

Assume that:

• N is an integer within the range [3..100,000];
• each element of array A is an integer within the range [−10,000..10,000].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

分别从左边和右边扫描数组，记录从左边以及右边到当前元素的最大连续区间和，然后再遍历一遍数组，找到left[i]+right[i]，别忘了再减掉当前元素，另外注意要把第一个元素和最后一个元素设为0。

 1 // you can use includes, for example:
 2 #include <algorithm>
 3 
 4 // you can write to stdout for debugging purposes, e.g.
 5 // cout << "this is a debug message" << endl;
 6 
 7 int solution(vector<int> &A) {
 8     // write your code in C++11
 9     if (A.size() <= 3) return 0;
10     vector<int> left(A), right(A);
11     int n = A.size();
12     left[0] = left[n-1] = 0;
13     right[0] = right[n-1] = 0;
14     for (int i = 1; i < n - 1; ++i) {
15         left[i] = max(left[i], left[i] + left[i-1]);
16         right[n-1-i] = max(right[n-1-i], right[n-1-i] + right[n-i]);
17     }
18     int res = 0;
19     for (int i = 1; i < n - 1; ++i) {
20         res = max(res, left[i] + right[i] - 2 * A[i]);
21     }
22     return res;
23 }

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