Sell Stock | Darren's Blog

Single time sell max profit
http://leetcode.com/2010/11/best-time-to-buy-and-sell-stock.html
Find i and j that maximizes Aj – Ai, where i < j.
To solve this problem efficiently, you would need to track the minimum value’s index. As you traverse, update the minimum value’s index when a new minimum is met. Then, compare the difference of the current element with the minimum value. Save the buy and sell time when the difference exceeds our maximum difference (also update the maximum difference).
void getBestTime(int stocks[], int sz, int &buy, int &sell) {
  int min = 0;
  int maxDiff = 0;
  buy = sell = 0;
  for (int i = 0; i < sz; i++) {
    if (stocks[i] < stocks[min])
      min = i;
    int diff = stocks[i] - stocks[min];
    if (diff > maxDiff) {
      buy = min;
      sell = i;
      maxDiff = diff;
    }
  }
}

http://blog.csdn.net/u013027996/article/details/19414967
You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times).
However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0){
            return 0;
        }
        int len = prices.length;
        int maxProfit = 0;
        for(int i = 1; i < len; i++){
            int tempProfit = prices[i] - prices[i-1];
            if(tempProfit > 0){
                maxProfit += tempProfit;
            }
        }
        return maxProfit;
    }

Best Time to Buy and Sell Stock III
You may complete at most two transactions.
we can use two arrays f[i] and b[i] to record the maximum profit for price[0...i−1] and price[i...n−1] , respectively. After that, we just need to find the maximum of f[i]+b[i].

public class BestTimeToBuyAndSellStockIII {

 

    // 392ms for 198 test cases

    public int maxProfit(int[] prices) {

        if (prices == null || prices.length == 0)

            return 0;

 

        // forward[i]: maximum profit for prices[0...i-1]; forward[0]=0

        int[] forward = new int[prices.length+1];

        int lowestBuyInPrice = prices[0];   // Lowest buy-in price up to now

        for (int i = 2; i <= prices.length; i++) {      // Traverse forwards

            forward[i] = Math.max(forward[i-1], prices[i-1]-lowestBuyInPrice);

            lowestBuyInPrice = Math.min(lowestBuyInPrice, prices[i-1]);

        }

 

        // backward[i]: maximum profit for prices[i...n-1]

        int[] backward = new int[prices.length];

        int highestSellOutPrice = prices[prices.length-1];   // Lowest buy-in price up to now

        for (int i = prices.length-2; i >= 0; i--) {    // Traverse backwards

            backward[i] = Math.max(backward[i+1], highestSellOutPrice-prices[i]);

            highestSellOutPrice = Math.max(highestSellOutPrice, prices[i]);

        }

 

        // Find the maximum of forward[i]+backward[i]

        int maximumProfit = 0;

        for (int i = 0; i < prices.length; i++) {

            maximumProfit = Math.max(maximumProfit, forward[i]+backward[i]);

        }

 

        return maximumProfit;

    }

}

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