[随缘一题]实现交易(fifo)

Posted1 by 呼延十 on February 24, 2019 Hot:

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根据先进先出原则实现交易. 例如:

buy 100 share(s) at $20 each
buy 20 share(s) at $24 each
buy 200 share(s) at $36 each
sell 150 share(s) at $30 each

得出计算结果 940.

优先卖掉持有时间最长的.

解题思路

直接使用Arraylist保存,卖出时从第一个开始即可.

当然也可以用队列做.

实现代码

/**
 * calculation the result
 * @param transactions
 * @return
 */
private Integer calculation(List<String> transactions) {
  int result = 0;

  //make the input to sell-100-20 format
  List<String> t = new ArrayList<>();
  for (String transaction : transactions) {
    if ("".equals(transaction)) {
      continue;
    }
    String[] ss = transaction.split(" ");
    t.add(ss[0] + "-" + ss[1] + "-" + ss[4].replace("$", ""));
  }

  for (int i = 0; i < t.size(); i++) {
    //cal while sell
    if (t.get(i).startsWith("sell")) {
      //get the num and the sell price
      int num = Integer.valueOf(t.get(i).split("-")[1]);
      int sellPrice = Integer.valueOf(t.get(i).split("-")[2]);
      //cal the buy before sell
      for (int j = 0; j < i; j++) {
        //sell shares, use FIFO.
        String[] sss = t.get(j).split("-");
        //if sell num < buy num, cal sell num shares in that transcation.
        if (num <= Integer.valueOf(sss[1])) {
          result += num * (sellPrice - Integer.valueOf(sss[2]));
          break;
        } else {
          //if sell num > buy num, cal all shares ,and cal new sellnum.
          result += Integer.valueOf(sss[1]) * (sellPrice - Integer.valueOf(sss[2]));
          num -= Integer.valueOf(sss[1]);
        }
      }
    }
  }

  return result;

}

完。




ChangeLog

2019-02-24 完成

以上皆为个人所思所得,如有错误欢迎评论区指正。

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