leetcode——賣股票系列

leetcode上賣股票系列題

只能交易一次

121. Best Time to Buy and Sell Stock

public int MaxProfit(int[] prices)
{
    if (prices == null || prices.Length == 0)
    {
        return 0;
    }
    var b = prices[0];
    var maxSoFar = 0;
    var maxcurr = 0;
    foreach (var p in prices)
    {
        maxcurr += p - b;
        maxcurr = Math.Max(maxcurr, 0);
        b = p;
        maxSoFar = Math.Max(maxcurr, maxSoFar);
    }
    return maxSoFar;
}

可以交易無限次

122. Best Time to Buy and Sell Stock II

public int MaxProfitII(int[] prices)
{
    if (prices == null || prices.Length == 0) return 0;
    var first = prices[0];
    var res = 0;
    for (var i = 1; i < prices.Length; i++)
    {
        var cur = prices[i] - first;
        first = prices[i];
        res += Math.Max(cur, 0);
    }
}
            

最多兩次交易

123. Best Time to Buy and Sell Stock III

public int MaxProfitIII(int[] prices)
{
    if (prices.Length == 0) return 0;
    int s1 = -prices[0], s2 = int.MinValue, s3 = int.MinValue, s4 = int.MinValue;

    for (int i = 1; i < prices.Length; ++i)
    {
        s1 = Math.Max(s1, -prices[i]);
        s2 = Math.Max(s2, s1 + prices[i]);
        s3 = Math.Max(s3, s2 - prices[i]);
        s4 = Math.Max(s4, s3 + prices[i]);
    }
    return Math.Max(0, s4);
}

最多k筆交易

188. Best Time to Buy and Sell Stock IV

public int MaxProfitIV(int k, int[] prices)
{
    var n = prices.Length;
    if (n <= 1) return 0;

    if (k >= n / 2)
    {//same as unlimited times
        var sum = 0;
        for (var i = 1; i < n; i++)
        {
            var cur = prices[i] - prices[i - 1];
            if (cur > 0) sum += cur;
        }
        return sum;
    }

    var dp = new int[k + 1, n];
    for (var i = 1; i < k + 1; i++)
    {
        var lastBuy = -prices[0];
        for (var j = 1; j < n; j++)
        {
            dp[i, j] = Math.Max(dp[i, j - 1], lastBuy + prices[j]);
            lastBuy = Math.Max(lastBuy, dp[i - 1, j] - prices[j]);
        }
    }
    return dp[k, n - 1];
}

不限次數有一天冷凍期

309. Best Time to Buy and Sell Stock with Cooldown

public int MaxProfitWithCoolDown(int[] p)
{
    //this is more easy to think
    if (p == null || p.Length < 2) return 0;
    var n = p.Length;
    var ownLast = -p[0];
    var notOwnLast = 0;
    var notOwnLast2 = 0;
    var res = 0;
    for (var i = 1; i < n; ++i)
    {
        var own = Math.Max(ownLast, notOwnLast2 - p[i]);
        var notOwn = Math.Max(ownLast + p[i], notOwnLast);

        notOwnLast2 = notOwnLast;
        ownLast = own;
        notOwnLast = notOwn;

        res = Math.Max(res, Math.Max(own, notOwn));
    }

    return res;
}

不限次數有手續費

714. Best Time to Buy and Sell Stock with Transaction Fee

public int MaxProfitWithFee(int[] prices, int fee)
{
    int cash = 0, hold = -prices[0];
    for (int i = 1; i < prices.Length; i++)
    {
        cash = Math.Max(cash, hold + prices[i] - fee);
        hold = Math.Max(hold, cash - prices[i]);
    }
    return cash;
}