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122.Best Time to Buy and Sell Stock II

📁 技术 ✏️ 更新于 2026年03月18日

与121不同的在于，121只能操作一次，而这个是可以操作任意次。

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Input: [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.

brute force

暴力搜索，没什么好说的

当前位置 $i$ ，搜索其后所有的可能答案，取最大的

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def brute_force(arr, start):
  if start >= len(arr):
    return 0
  max_profit = 0
  for i in range(start, len(arr)):
    tmp_max_profit = 0
    for j in range(start+1, len(arr)):
      if arr[j] > arr[i]:
        profit = brute_force(arr, j+1) + arr[j] - arr[i]
        if profit > tmp_max_profit:
          tmp_max_profit = profit
    if tmp_max_profit>max_profit:
      max_profit = tmp_max_profit
  return max_profit

时间复杂度$O(n^n)$ ， 空间复杂度$O(n)$

Peak Valley Approach

给定的price数组为$[7, 1, 5, 3, 6, 4]$. 绘制图片有（来自leetcode）

显然有，最终的收益来自于所有的峰值减去谷值之和

$Total_Profit = \sum_i(height(peak_i)-height(valley_i))$

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def Peak_Valley(prices):
  valley = prices[0]
  peak  = prices[0]
  idx = 0
  max_profit = 0
  while idx < len(prices)-1:
    while idx < len(prices)-1 and prices[idx+1] <= prices[idx]:
      idx+=1
    vally = prices[idx]
    while idx < len(prices)-1 and prices[idx+1] > prices[idx]:
      idx+=1
    peak = prices[idx]
    max_profit += peak-vally
  return max_profit

时间复杂度$O(n)$, 空间复杂度$O(1)$

Simple One Pass

跟上面略微不同的是，只要斜率是正的，就一直买入卖出就可以获得最大利润

Profit Graph

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def One_Pass(prices):
  max_profit = 0
  for idx, price in enumerate(prices):
    if idx > 0 and price > prices[idx-1]:
      max_profit += price-prices[idx-1]
  return max_profit

时间复杂度$O(n)$, 空间复杂度$O(1)$