Best Time to Buy and Sell Stock

leetcode  123. Best Time to Buy and Sell Stock III

下面这种方法构思巧 难想到

既是dp（组合，先卖后买，先买后卖，需要比较）  又是贪心（每一单个操作最优，即是全局最优）

public class Solution {    public int maxProfit(int[] prices) {        // these four variables represent your profit after executing corresponding transaction        // in the beginning, your profit is 0.         // when you buy a stock ,the profit will be deducted of the price of stock.        int firstBuy = Integer.MIN_VALUE, firstSell = 0;        int secondBuy = Integer.MIN_VALUE, secondSell = 0;        for (int curPrice : prices) {            if (firstBuy < -curPrice) firstBuy = -curPrice; // the max profit after you buy first stock            if (firstSell < firstBuy + curPrice) firstSell = firstBuy + curPrice; // the max profit after you sell it            if (secondBuy < firstSell - curPrice) secondBuy = firstSell - curPrice; // the max profit after you buy the second stock            if (secondSell < secondBuy + curPrice) secondSell = secondBuy + curPrice; // the max profit after you sell the second stock        }        return secondSell; // secondSell will be the max profit after passing the prices    }}

下面这种方法 dp  基本上绝大部分问题都可以化为dp问题

class Solution {public:    int maxProfit(vector<int> &prices) {        // f[k, ii] represents the max profit up until prices[ii] (Note: NOT ending with prices[ii]) using at most k transactions.         // f[k, ii] = max(f[k, ii-1], prices[ii] - prices[jj] + f[k-1, jj]) { jj in range of [0, ii-1] }        //          = max(f[k, ii-1], prices[ii] + max(f[k-1, jj] - prices[jj]))        // f[0, ii] = 0; 0 times transation makes 0 profit        // f[k, 0] = 0; if there is only one price data point you can't make any money no matter how many times you can trade        if (prices.size() <= 1) return 0;        else {            int K = 2; // number of max transation allowed            int maxProf = 0;            vector<vector<int>> f(K+1, vector<int>(prices.size(), 0));            for (int kk = 1; kk <= K; kk++) {                int tmpMax = f[kk-1][0] - prices[0];                for (int ii = 1; ii < prices.size(); ii++) {                    f[kk][ii] = max(f[kk][ii-1], prices[ii] + tmpMax);                    tmpMax = max(tmpMax, f[kk-1][ii] - prices[ii]);                    maxProf = max(f[kk][ii], maxProf);                }            }            return maxProf;        }    }};

leetcode 很好的一点就是将一类题目放在一起，好下面我们开始解决问题：

贪心算法： https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/#/description

public class Solution {    public int maxProfit(int[] prices) {        if(prices==null) return 0;        int len = prices.length;        if(len==1) return 0;        int i = 0;        int j = 1;        int res = 0;        while(j<=len-1){            if(prices[j]>prices[i])              {                res += prices[j]-prices[i];                          }            i = j;            j = j +1;        }        return res;            }}