【leetcode】Best Time to Buy and Sell Stock

最新推荐文章于 2023-02-28 14:37:52 发布

原创最新推荐文章于 2023-02-28 14:37:52 发布 · 291 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #算法 #c

算法专栏收录该内容

21 篇文章

订阅专栏

本文介绍了一种寻找股票交易最大利润的算法。该算法通过计算每日价格差值并确定最佳买入卖出时机来实现最大收益。核心思想在于追踪连续正差值以累积最大利润。

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Accept: 5ms

思路：

先两两相减得到前后2天的差值：diff[i] = prices[i+1] - prices[i];
问题就简化为在diff数组中找到和最大的连续子串问题：当前为i，总和为sum，下一个为i+1，则：
- 如果 diff[i+1] >=0，则sum += diff[i+1];
- 如果 diff[i+1] < 0，则：
  - 如果 sum + diff[i+1] > 0，则选择i+1，
  - 否则：在i+1这里断开。

int diff[100000];

int maxProfit(int prices[], int n){
  for (int i = 0; i < n - 1; ++i) {
    diff[i] = prices[i+1] - prices[i];
  }

  int i = 0;
  for (; i < n - 1 && diff[i] <= 0; ++i) /* nothing */;

  int max = 0;
  int sum = 0;
  for (; i < n - 1; ++i) {
    int next = diff[i] + sum;
    if (next <= 0) {
      sum = 0;
      continue;
    }
    sum += diff[i];
    if (sum > max) {
      max = sum;
    }
  } // for

  return max;
}