Best Time to Buy and Sell Stock III
阿新 • • 發佈:2017-08-13
urn 曾經 -s 查找 tip 依據 都是 sign snippet
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 two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
想不出來,網上搜了一個算法,從頭算一遍。然後在從尾算一遍當前最大值,然後從不同的i切割。前後加起來最大的情況。 另外還看到了一個求最大k次交易的一個通用算法,是依據一篇論文寫出來的,看來學術確實能用到尋常的編程裏來,事實上想想如今的二分查找等算法不都是曾經的論文麽。
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
想不出來,網上搜了一個算法,從頭算一遍。然後在從尾算一遍當前最大值,然後從不同的i切割。前後加起來最大的情況。
int maxProfitIII(vector<int> &prices) { int len = prices.size(); if (len == 0) return 0; vector<int> history(len, 0); vector<int> future(len, 0); int low = prices[0]; for (int i = 1; i < len; i++) { history[i] = max(history[i-1], prices[i] - low); low = min(low, prices[i]); } int high = prices[len-1]; for (int i = len-2; i >= 0; i--) { future[i] = max(future[i+1], high - prices[i]); high = max(high, prices[i]); } int maxProfit = 0; // 網上說第一次的賣和第二次的買能夠在同一次。所以直接都是i相加就能夠了 for (int i = 0; i < len; i++) { maxProfit = max(maxProfit, history[i] + future[i]); } return maxProfit; }
Best Time to Buy and Sell Stock III