LeetCode

这道题我一开始想到用递归方法，可以把规模大的问题变成规模小的问题，但是觉得递归的时间复杂度很高，因为它会把相同的问题进行重复计算，然后我想是不是有什么down-up的方法，先把所有的子问题的结果保存起来，但是发现问题的最优解并不能由子问题的最优解推导出来。最后就想到买股票的时候，我们在一个局部极小的时候买进，然后在一个局部极大的时候卖出，这样就可以通过多次买进卖出交易获得最大收益。接下来的问题就是，确定什么时候是极小值，什么时候是极大值。

下面是AC代码：

    /**
     * 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).
     * @param prices
     * @return
     */
    public int maxProfit2(int[] prices){
        if(prices==null || prices.length <= 1)
            return 0;
        int profit = 0;
        int len = prices.length;
        //the local minimum is marked as -1;the local maximum is marked as 1; otherwise is 0
        int[] mark = new int[len];
        int[] left = new int[len];//mark whether the left is <= or >=
        int[] right = new int[len];//mark whether the right is <= or >=
        // the first number is initialized as 0
        left[0] = 0;
        //get the left array, by one traversal
        int i=1;
        while(i<len){
            if(prices[i-1]<prices[i])
                left[i] = -1;
            else if(prices[i-1]>prices[i])
                left[i] = 1;
            else
                left[i] = left[i-1];
            i++;
        }
        //initialize the last element 
        right[len-1] = 0;
        i = len-2;
        while(i>=0){
            if(prices[i+1]<prices[i])
                right[i] = -1;
            else if(prices[i+1]>prices[i])
                right[i] = 1;
            else
                right[i] = right[i+1];
            i--;
        }
        //get the mark
        i = 0;
        while(i<len){
            if(left[i] == right[i]){
                mark[i] = left[i]*(-1);
            }else {
                if(left[i] == 0){
                    mark[i] = right[i]*(-1);
                }
                else if(right[i] == 0)
                        mark[i] = left[i]*(-1);
                else
                    mark[i] = 0;
            }
            i++;
        }
        int buy=0 ,sell = 0;
        boolean f1 = false, f2 = false;
        for(i=0;i<len;i++){
            if(f1==false && mark[i] == -1)//at the minimum time buy.
            {
                buy = prices[i];
                f1 = true;
            }
            if(f2 == false && f1 == true && mark[i] == 1)//at the maximum time sell, by the condition that we have bought the stock;
            {
                sell = prices[i];
                f2 = true;
            }
            if(f1 && f2)
            {
                profit += (sell - buy); 
                f1 = false;
                f2 = false;
            }
        }
        return profit;
     }