LeetCode OJ - Best Time to Buy and Sell Stock III
2014-06-26 15:39
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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).
考虑数组[1,2,4,2,5,7,2,4,9,0]
分析:考虑了半天终于有思路了,类似于Best Time to Buy and Sell Stock I,将数字分为两个区间[0, i]和[i+1 , n-1],分别求出两个区域的最大收益并相加即可,用迭代求出最后的收益。
方法一:i 取值选在每次的波峰位置,接着在[i+1 , n-1]中找出最大收益,时间复杂度为O(n^2)
方法二:对于每一个i求出 [0, i]最大收益front[i], [i+1, n-1]最大收益back[i], 求出max{font[i] + back[i]},这里有一些无效计算但是时间复杂度为O(n)。
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).
考虑数组[1,2,4,2,5,7,2,4,9,0]
分析:考虑了半天终于有思路了,类似于Best Time to Buy and Sell Stock I,将数字分为两个区间[0, i]和[i+1 , n-1],分别求出两个区域的最大收益并相加即可,用迭代求出最后的收益。
方法一:i 取值选在每次的波峰位置,接着在[i+1 , n-1]中找出最大收益,时间复杂度为O(n^2)
方法二:对于每一个i求出 [0, i]最大收益front[i], [i+1, n-1]最大收益back[i], 求出max{font[i] + back[i]},这里有一些无效计算但是时间复杂度为O(n)。
class Solution { public: int maxProfit(vector<int> &prices) { int len = prices.size(); if(len < 2) return 0; int *front = new int[len]; int *back = new int[len]; memset(front, 0, len * sizeof(int)); memset(back, 0, len * sizeof(int)); int minprice = prices[0]; for(int i = 1; i < len; i++) { front[i] = prices[i] - minprice; minprice = min(minprice, prices[i]); } int maxprice = prices[len - 1]; for(int i = len - 2; i >= 0; i--) { back[i] = max(back[i+1], maxprice - prices[i+1]); //back[i+1]代表历史最大 maxprice = max(maxprice, prices[i+1]); } int ret = 0; for(int i = 0; i < len; i++) { ret = max(ret, front[i] + back[i]); } return ret; } };
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