Codility-MaxProfit

最新推荐文章于 2021-02-21 17:05:28 发布

原创最新推荐文章于 2021-02-21 17:05:28 发布 · 948 阅读

0 ·

CC 4.0 BY-SA版权

Codility 同时被 2 个专栏收录

12 篇文章

订阅专栏

MaxProfit

1 篇文章

订阅专栏

本文介绍了一种计算给定时间段内股票交易最大可能利润的算法。通过一次遍历数组，找到买入和卖出的最佳时机，实现O(N)的时间复杂度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Task description

A zero-indexed array A consisting of N integers is given. It contains daily prices of a stock share for a period of N consecutive days. If a single share was bought on day P and sold on day Q, where 0 ≤ P ≤ Q < N, then the profit of such transaction is equal to A[Q] − A[P], provided that A[Q] ≥ A[P]. Otherwise, the transaction brings loss of A[P] − A[Q].

For example, consider the following array A consisting of six elements such that:

  A[0] = 23171  
  A[1] = 21011  
  A[2] = 21123
  A[3] = 21366  
  A[4] = 21013  
  A[5] = 21367

If a share was bought on day 0 and sold on day 2, a loss of 2048 would occur because A[2] − A[0] = 21123 − 23171 = −2048. If a share was bought on day 4 and sold on day 5, a profit of 354 would occur because A[5] − A[4] = 21367 − 21013 = 354. Maximum possible profit was 356. It would occur if a share was bought on day 1 and sold on day 5.

Write a function,

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

that, given a zero-indexed array A consisting of N integers containing daily prices of a stock share for a period of N consecutive days, returns the maximum possible profit from one transaction during this period. The function should return 0 if it was impossible to gain any profit.

For example, given array A consisting of six elements such that:

  A[0] = 23171  
  A[1] = 21011  
  A[2] = 21123
  A[3] = 21366  
  A[4] = 21013  
  A[5] = 21367

the function should return 356, as explained above.

Assume that:

N is an integer within the range [0..400,000];
each element of array A is an integer within the range [0..200,000].

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

此题事实上就是给定一个数组A[0]...A[n-1]，求最大的profit=A[q]-A[p]，其中0<=p<q<n。若数组为空或profit为负值则返回0。

很容易想到O(n^2)的暴力解法。事实上，对于A[q]-A[p]，若给定q，则其最大值在A[p]最小处取得。令max_ending表示给定q值时profit的最大值（用tmp表示A[p]的最小值），则对q从1~n-1进行一次遍历，在遍历时，若max_ending的值比原来profit值更大，则将profit更新为max_ending，否则不变。这样，一次遍历下来profit就更新为A[q]-A[p]的最大值了。

// 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 profit = 0;
    int tmp = A[0];
    int max_ending = 0;
    for(unsigned int i=1;iA[i])
        {
            tmp = A[i];
        }
    }
    return profit;
}