Codility-MaxSliceSum&MaxDoubleSlice

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Codility 同时被 3 个专栏收录

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MaxSlice

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本文深入解析了MaxSliceSum和MaxDoubleSlice算法，详细阐述了这两种算法的实现逻辑、核心思想及应用背景。通过具体代码示例，读者能够清晰地理解如何在给定数组中寻找最大子数组和最大双子数组和，适用于计算机科学领域的算法学习与实践。

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MaxSliceSum:

Task description

A non-empty zero-indexed array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The sum of a slice (P, Q) is the total of A[P] + A[P+1] + ... + A[Q].

Write a function:

int solution(const vector<int> &A);

that, given an array A consisting of N integers, returns the maximum sum of any slice of A.

此题的思路就是设置一个max_ending，记录右边界固定为i时的最大部分和。max_slice不断更新为max_ending与max_slice中较大者。

代码：

// you can use includes, for example:
 #include 

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(const vector &A) {
    // write your code in C++11
    if(A.empty())
    return 0;
    int max_ending = A[0];
    int max_slice = A[0];
    for(unsigned int i=1;i




MaxDoubleSlice:



Task description


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(vector<int> &A);












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.

此题是maxslice的进阶题。事实上就是求A[x]+...+A[z]，然后减去A[x],A[y],A[z]后的部分和。和maxsilice的思路一样，我们仍然是通过枚举右边界来更新maxdoubleslice。但有三个点x,y,z，只枚举右边界就能行吗？是的，因为当右边界从i变成i+1时：1、若之前的max_ending为负值，则新的max_ending直接更新成A[i]（可找出xyz使得max_ending=A[i]）2、若之前的max_ending非负，但A[i]小于A[y]，则将y更新成i（y处必须为最小值，因A[y]不会被加入部分和中），max_ending更新成以i为右边界的最大部分和（max_left）；否则保持y不变，max_ending更新成max_ending+A[i]。
当左边序列的右边界为i时，max_left更新成max_left+A[i-1]和A[i-1]中较大者。
每次遍历都将max_slice更新为原max_slice和max_ending中较大者，则遍历完成后得到最优解。
代码：

// you can use includes, for example:
 #include 

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
inline int max(int a,int b,int c)
{
    if(a &A) {
    // write your code in C++11
    if(A.size()<4)
    return 0;
    //右边界为i时两段最大值max_ending
    //中间点为i时左子列最大部分和max_left
    //整个数组最大部分和max_slice
    int max_ending = 0 , max_left = 0 , max_slice = 0;
    for(unsigned int i=3;i