Viiisit [LeetCode] - 122. Best Time to Buy and Sell Stock II

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Problem - Best Time to Buy and Sell Stock II

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

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

翻譯蒟蒻

找到在一系列天數中，以買進和賣出股票的方式獲得的最大利潤。
限制是一次只能持有一股，但可以在同一天內先買進再賣出。

• Example 1:

  Input: prices = [7,1,5,3,6,4]
  Output: 7
  Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
  Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
  Total profit is 4 + 3 = 7.

• Example 2:

  Input: prices = [1,2,3,4,5]
  Output: 4
  Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
  Total profit is 4.

• Example 3:

  Input: prices = [7,6,4,3,1]
  Output: 0
  Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

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Solution - JavaScript

Solution 1:

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
  let maxProfit = 0;
  for (let i = 1; i < prices.length; i++) {
    if (prices[i] > prices[i - 1]) {
      maxProfit += prices[i] - prices[i - 1]
    }
  }
  return maxProfit
};

• 這裡用 if 去判斷後面的數如果大於前一個就累加進 maxProfit 。

Solution 2:

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

• 用 Math.max 方法跟 0 去比較，把相減為負數的去除。

Solution - Ruby

def max_profit(prices)
  max_profit = 0

  (1...prices.length).each do |i|
    max_profit += [0, prices[i] - prices[i-1]].max
  end
  
  max_profit
end

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