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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 a double slice.
The sum of 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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原文地址：http://www.cnblogs.com/easonliu/p/4453314.html
