1.Best Time to Buy and Sell Stock

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.

題目的意思是：

思路：動態規劃法，要想獲得最大收益，就得在最低的價格時候買入，在最高的價格時候賣出。

O(n)時間，O(1)空間。

/**
     * 從前向後遍歷陣列prices，記錄當前出現過的最低價格low，作為買
     * 入價格，並計算以當天價格出售的收益，作為可能的最大收益res，
     * 整個遍歷過程中，出現過的最大收益(即res最大)就是所求。
    */
    public int maxProfit(int[] prices) {
        if(prices==null||prices.length==0){
            return 0;
        }
        
        int low=prices[0];//記錄最小值
        int res=0;//記錄最大差
        for(int i=1;i<prices.length;i++){
            if(low>prices[i]){
                low=prices[i];
            }
            else if(prices[i]-low>res){
                res=prices[i]-low;
            }
        }
        return res;
    }
 

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，陣列的第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。

思路：貪心法。從前向後遍歷陣列prices，只要當天的價格高於前一天的價格，就算入收益。

程式碼：時間O(n)，空間O(1)。

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

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，陣列的第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：動態規劃法。以第i天為分界線，計算第i天之前進行一次交易的最大收益preProfit[i]，和第i天之後進行一次交易的最大收益postProfit[i]，最後遍歷一遍，max{preProfit[i] + postProfit[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。

程式碼：時間O(n)，空間O(n)。

public int maxProfit(int[] prices) {
        if(prices==null||prices.length==0){
            return 0;
        }
        int n=prices.length;
        int[] preProfit=new int[n];//記錄第i天以前進行一次買賣的最大收益；
        int[] postProfit=new int[n];//記錄第i天以後進行一次買賣的最大收益；
        
        int curMin=prices[0]; 
        for(int i=1;i<n;i++){
            curMin=curMin>prices[i]?prices[i]:curMin;
            preProfit[i]=(prices[i]-curMin)>preProfit[i-1]?prices[i]-curMin:preProfit[i-1];
        }
        
        int curMax=prices[n-1];
        for(int i=n-2;i>=0;i--){
            curMax=curMax>prices[i]?curMax:prices[i];
            postProfit[i]=postProfit[i+1]>curMax-prices[i]?postProfit[i+1]:curMax-prices[i];
        }
        
        int maxProfit=0;
        for(int i=1;i<n;i++){
            maxProfit=maxProfit>(postProfit[i]+preProfit[i])?maxProfit:(postProfit[i]+preProfit[i]);
        }
        return maxProfit;
    }