// Source : https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ // Author : Hao Chen // Date : 2014-06-18 /***************************************************************************************************** * * 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 * (i.e., buy one and sell one share of the stock multiple times). * * Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock * before you buy again). * * Example 1: * * Input: [7,1,5,3,6,4] * Output: 7 * Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. * Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3. * * Example 2: * * Input: [1,2,3,4,5] * Output: 4 * Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. * Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are * engaging multiple transactions at the same time. You must sell before buying again. * * Example 3: * * Input: [7,6,4,3,1] * Output: 0 * Explanation: In this case, no transaction is done, i.e. max profit = 0. ******************************************************************************************************/ class Solution { public: int maxProfit(vector<int>& prices) { return maxProfit02(prices); return maxProfit01(prices); } // Solution 1 // find all of ranges: which start a valley with the nearest peak after // add their delta together // int maxProfit01(vector<int> &prices) { int max = 0; int low = -1; int len = prices.size(); for (int i=0; i < len - 1; i++){ //meet the valley, then goes up if (prices[i] < prices[i+1] && low < 0 ) { low = i; } //meet the peak, then goes down if (prices[i] > prices[i+1] && low >= 0) { max += ( prices[i] - prices[low] ) ; low = -1; // reset the `low` } } // edge case if ( low >= 0 ) { max += ( prices[prices.size()-1] - prices[low] ); } return max; } // Solution 2 // if we find we can earn money, we just sell int maxProfit02(vector<int>& prices) { int profit = 0 ; for(int i=1; i< prices.size(); i++) { profit += max(0, prices[i] - prices[i-1]); } return profit; } };