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; } }
