121. 买卖股票的最佳时机(Best Time to Buy and Sell Stock)E

英文题目

• You are given an array prices where prices[i] is the price of a given stock on the ith day.

• You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

• Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

• Example 1:

• Input: prices = [7,1,5,3,6,4]
  Output: 5
  Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
  Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
  

• Example 2:

• Input: prices = [7,6,4,3,1]
  Output: 0
  Explanation: In this case, no transactions are done and the max profit = 0.
  

• Constraints:

• 1 <= prices.length <= 10^5

• 0 <= prices[i] <= 10^4

中文题目

• 给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。

• 如果你最多只允许完成一笔交易（即买入和卖出一支股票一次），设计一个算法来计算你所能获取的最大利润。

• 注意：你不能在买入股票前卖出股票。

• 示例 1:

• 输入: [7,1,5,3,6,4]
  输出: 5
  解释: 在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。
  注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格；同时，你不能在买入前卖出股票。
  

• 示例 2:

• 输入: [7,6,4,3,1]
  输出: 0
  解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。
  

暴力解法

• 找出给定两个数字之间的最大差值(即最大利润)

• 注意，以下例子会超时

• 时间复杂度：O(n^2)，空间复杂度O(1)

  # python: 超时
  class Solution:
      def maxProfit(self, prices: List[int]) -> int:
          res = 0
          # 遍历所有的买入价格，最后一天买入无用
          for buy in range(len(prices)-1):
              # 遍历买入价格后一天的所有卖出价格
              for sell in range(buy+1, len(prices)):
                  res = max(res, prices[sell] - prices[buy])
          return res
  

  // c++: 超时
  class Solution {
  public:
      int maxProfit(vector<int>& prices) {
          int res = 0;
          for (int buy = 0; buy < prices.size()-1; ++buy) {
              for (int sell = buy+1; sell < prices.size(); ++sell) {
                  res = max(res, prices[sell] - prices[buy]);
              }
          }
          return res;
      }
  };
  

  // java: 超时
  class Solution {
      public int maxProfit(int[] prices) {
          int res = 0;
          for (int buy = 0; buy < prices.length - 1; ++buy) {
              for (int sell = buy+1; sell < prices.length; ++sell) {
                  res = Math.max(res, prices[sell] - prices[buy]);
              }
          }
          return res;
      }
  }
  

  // go: 超时
  func maxProfit(prices []int) int {
      res := 0
      for buy := 0; buy < len(prices) - 1; buy++ {
          for sell := buy+1; sell < len(prices); sell++ {
              res = maxInt(res, prices[sell]-prices[buy])
          }
      }
      return res
  }
  
  func maxInt(a, b int) int {
      if a >= b {
          return a
      } else {
          return b
      }
  }
  

  // javascript: 超时
  /**
   * @param {number[]} prices
   * @return {number}
   */
  var maxProfit = function(prices) {
      let res = 0;
      for (let buy = 0; buy < prices.length - 1; ++buy) {
          for (let sell = buy+1; sell < prices.length; ++sell) {
              res = Math.max(res, prices[sell] - prices[buy]);
          }
      }
      return res;
  };
  

一次遍历(贪心)

• 一次遍历中，记录历史最低点，在每一天都进行计算，我今天卖出能赚多少钱

• 时间复杂度O(n)，空间复杂度O(1)

  # python: 时间 228 ms, 击败 51.53%; 内存 25.9 MB, 击败 61.17%
  class Solution:
      def maxProfit(self, prices: List[int]) -> int:
          # 初始化时，假设第一天为历史最低点
          minPrice, res = prices[0], 0
          for i in range(1, len(prices)):
              res = max(res, prices[i] - minPrice)
              minPrice = min(minPrice, prices[i])
          return res
  

  // c++: 时间 108 ms, 击败 55.45%; 内存 91.1 MB, 击败 77.38%
  class Solution {
  public:
      int maxProfit(vector<int>& prices) {
          int minPrice = prices[0], res = 0;
          for (int i = 1; i < prices.size(); ++i) {
              res = max(res, prices[i] - minPrice);
              minPrice = min(minPrice, prices[i]);
          }
          return res;
      }
  };
  

  // java: 时间 2 ms, 击败 43.37%; 内存 57.4 MB, 击败 75.6%
  class Solution {
      public int maxProfit(int[] prices) {
          int minPrice = prices[0], res = 0;
          for (int i = 1; i < prices.length; ++i) {
              res = Math.max(res, prices[i] - minPrice);
              minPrice = Math.min(minPrice, prices[i]);
          }
          return res;
      }
  }
  

  // go: 时间 104 ms, 击败 89.46%; 内存 8.05 MB, 击败 32.31%
  func maxProfit(prices []int) int {
      minPrice, res := prices[0], 0
      for i := 0; i < len(prices); i++ {
          res = maxInt(res, prices[i] - minPrice)
          minPrice = minInt(minPrice, prices[i])
      }
      return res
  }
  
  func maxInt(a, b int) int {
      if a > b {
          return a
      } else {
          return b
      }
  }
  
  func minInt(a, b int) int {
      if a < b {
          return a
      } else {
          return b
      }
  }
  

  // javascript: 时间 68 ms, 击败 97.71%; 内存 49.34 MB, 击败 75.73%
  /**
   * @param {number[]} prices
   * @return {number}
   */
  var maxProfit = function(prices) {
      let minPrice = prices[0], res = 0;
      for (let i = 0; i < prices.length; ++i) {
          res = Math.max(res, prices[i] - minPrice);
          minPrice = Math.min(minPrice, prices[i]);
      }
      return res;
  };