買賣股票 I II III

Best Time to Buy and Sell Stock I

Description: Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find
the maximum profit.

題意：用一個數組表示股票每天的價格，數組的第i個數表示股票在第i天的價格。 如果只允許進行一次交易，也就是說只允許買一支股票並賣掉，求最大的收益。

分析：從前向後遍歷數組，記錄當前出現過的最低價格，作爲買入價格，並計算以當天價格出售的收益，作爲可能的最大收益，整個遍歷過程中，出現過的最大收益就是所求。

public int maxProfit(int[] prices) {
        int max = 0, min = Integer.MAX_VALUE;
        int diff;
        for(int i=0;i<prices.length;i++){
            if(prices[i] < min)
                min = prices[i];
            diff = prices[i] - min;
            if(diff > max)
                max = diff;
        }
        return max;
    }

Best Time to Buy and Sell Stock II

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 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).

題目：用一個數組表示股票每天的價格，數組的第i個數表示股票在第i天的價格。交易次數不限，但一次只能交易一支股票，也就是說手上最多隻能持有一支股票，求最大收益。

分析：貪心法。從前向後遍歷數組，只要當天的價格高於前一天的價格，就算入收益。

public int maxProfit(int[] prices) {
        int sum = 0;
        for(int i=1;i<prices.length;i++){
            if(prices[i] - prices[i-1] > 0)
                sum += prices[i] - prices[i-1];
        }
        return sum;
    }

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 (ie, you must sell the stock before you buy again).

題意：用一個數組表示股票每天的價格，數組的第i個數表示股票在第i天的價格。最多交易兩次，手上最多隻能持有一支股票，求最大收益。

分析：以第i天爲分界線，計算第i天之前進行一次交易的最大收益maxFromLeft[i]，和第i天之後進行一次交易的最大收益maxFromRight[i]，最後遍歷一遍，max{maxFromLeft[i] + maxFromRight[i]} (0≤i≤n-1)就是最大收益。第i天之前和第i天之後進行一次的最大收益求法同Best Time to Buy and Sell Stock I。

public int maxProfit(int[] prices) {
        if (prices.length <= 1)
            return 0;
        if (prices.length == 2)
            return prices[1] > prices[0] ? prices[1] - prices[0] : 0;
        int max = Integer.MIN_VALUE;
        int diff;

        int[] maxFromLeft = new int[prices.length];
        int minL = Integer.MAX_VALUE;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < minL)
                minL = prices[i];
            diff = prices[i] - minL;
            if (diff > max)
                max = diff;
            maxFromLeft[i] = max;
        }

        int maxR = Integer.MIN_VALUE;
        max = Integer.MIN_VALUE;
        int[] maxFromRight = new int[prices.length];
        for (int i = prices.length - 1; i >= 0; i--) {
            if (prices[i] > maxR)
                maxR = prices[i];
            diff = maxR - prices[i];
            if (diff > max)
                max = diff;
            maxFromRight[i] = max;
        }
        max = Integer.MIN_VALUE;
        for (int i = 0; i < prices.length; i++) {
            diff = maxFromLeft[i] + maxFromRight[i];
            if (diff > max)
                max = diff;
        }
        return max;
    }