Best Time to Buy and Sell Stock1,2,3，4

找到最低值和最高值

int maxProfit(vector<int>& prices) {
        if(prices.size()<)return ;
        int profit=;
        int cur_min=prices[];
        for(int i=;i<prices.size();i++)
        {
            profit=max(profit,prices[i]-cur_min);//记录最大利润
            cur_min=min(cur_min,prices[i]);//保留购买最小值
        }
        return profit;
    }

2、

计算差分序列，大于0加入

int maxProfit(vector<int>& prices) {
        if(prices.size()<)return ;
        int profit=;
        int diff=;
        for(int i=;i<prices.size();i++)
        {
            diff=prices[i]-prices[i-];
            if(diff>)profit+=diff;
        }
        return profit;
    }

3、

把交易分成两次，分别完成，最后将利润相加求最大。

public int maxProfit(int[] prices) {
        if (prices.length < ) return ;

        int n = prices.length;
        int[] preProfit = new int[n];
        int[] postProfit = new int[n];

        int curMin = prices[];
        for (int i = ; i < n; i++) {
            curMin = Math.min(curMin, prices[i]);
            preProfit[i] = Math.max(preProfit[i - ], prices[i] - curMin);
        }

        int curMax = prices[n - ];
        for (int i = n - ; i >= ; i--) {
            curMax = Math.max(curMax, prices[i]);
            postProfit[i] = Math.max(postProfit[i + ], curMax - prices[i]);
        }

        int maxProfit = ;
        for (int i = ; i < n; i++) {
            maxProfit = Math.max(maxProfit, preProfit[i] + postProfit[i]);
        }

        return  maxProfit;
    }

4、

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 k transactions. Note: 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 k, int[] prices) {
        if (prices.length < ) return ;

        int days = prices.length;
        if (k >= days) return maxProfit2(prices);

        int[][] local = new int[days][k + ];
        int[][] global = new int[days][k + ];

        for (int i = ; i < days ; i++) {
            int diff = prices[i] - prices[i - ];

            for (int j = ; j <= k; j++) {
                local[i][j] = Math.max(global[i - ][j - ], local[i - ][j] + diff);
                global[i][j] = Math.max(global[i - ][j], local[i][j]);
             }
        }

        return global[days - ][k];
    }

    public int maxProfit2(int[] prices) {
        int maxProfit = ;

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

        return maxProfit;
    }

参考：http://liangjiabin.com/blog/2015/04/leetcode-best-time-to-buy-and-sell-stock.html

可以交易k次，没看懂，感觉自己好笨。。。