123. Best Time to Buy and Sell Stock III

Description

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

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [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.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Tags: Math, String

题意

给你一个数组,第 ii 个元素表示第 ii 天股票的价格。 你最多可以交易两次。 请设计一个算法,求最大收益。

注意:必须先买再卖,且每天只能买或者卖一次。

题解

思路1

在整个区间的每一点切开, 然后分别计算左子区间和右子区间的最大值,然后再用O(n)时间找到整个区间的最大值。

  • 遍历一遍数组,求[0,i−1][0,i−1]区间的最大利润f(i),具体做法是找当前最低价格low,判断是要以low买入当天卖出,还是不动
  • 从后往前遍历,求[i,n−1][i,n−1]区间的最大利润g(i),具体做法是找当前最高价格high,判断是要当天买入以high卖出,还是不动
  • 遍历,求最大利润max(f(i)+g(i))
func maxProfit(prices []int) int {
    t1_b, t1_s, t2_b, t2_s := math.MinInt32, 0, math.MinInt32, 0

    for _, v := range prices {
        t1_b = max(t1_b, 0-v)
        t1_s = max(t1_s, t1_b+v)
        t2_b = max(t2_b, t1_s-v)
        t2_s = max(t2_s, t2_b+v)
    }
    return t2_s
}
func max(a, b int) int {
    if a < b {
        return b
    }
    return a
}

结语

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