Best Time to Buy and Sell Stock II
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).
不要想太多，就是整个波动中所有上升曲线的和。
1 class Solution { 2 public: 3 /* 大数据集妥妥的超时的暴力方法，啊！太傻了！捉急啊！！！ 4 int result; 5 6 void getMaxProfit(int buy_price, int day_start, int profit_before, const vector & prices) { 7 if (day_start == prices().size()) { 8 if (profit_before > result) { 9 result = profit_before;10 return;11 }12 }13 for (int i = day_start; i < prices.size(); ++i) {14 if (prices[i] < buy_price) {15 getMaxProfit(prices[i], i + 1, profit_before, prices); // 买16 getMaxProfit(buy_price, i + 1, profit_before, prices); // 不买17 } else {18 getMaxProfit(INT_MAX, i + 1, profit_before + prices[i] - buy_price, prices); // 卖19 getMaxProfit(buy_price, i + 1, profit_before, prices); // 不卖20 }21 }22 }23 24 int maxProfit(vector &prices) {25 // Start typing your C/C++ solution below26 // DO NOT write int main() function27 result = 0;28 getMaxProfit(INT_MAX, 0, 0, prices);29 return result;30 }31 */32 33 int maxProfit(vector &prices) {34 // Start typing your C/C++ solution below35 // DO NOT write int main() function36 // 所有上升线段长度和......37 if (0 == prices.size()) {38 return 0;39 }40 int max_profit = 0, last_min = prices.at(0);41 for (int i = 1; i < prices.size(); ++i) {42 if (prices.at(i) < prices.at(i - 1)) {43 max_profit += (prices.at(i - 1) - last_min);44 last_min = prices.at(i);45 }46 }47 max_profit += (prices.at(prices.size() - 1) - last_min);48 return max_profit;49 }50 };