【LeetCode】Best Time to Buy and Sell Stock & Best Time to Buy and Sell Stock
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参考链接
http://www.cnblogs.com/wdpp/archive/2011/12/28/2386704.html
http://hi.baidu.com/liu_a_meng/item/e2dbd8e4ac11a0c4baf37d4f
http://www.cnblogs.com/remlostime/archive/2012/11/06/2757434.html
II
题目描述
Best Time to Buy and Sell Stock
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
题目分析
总结
代码示例
class Solution {public: int maxProfit(vector<int> &prices) { // Start typing your C/C++ solution below // DO NOT write int main() function if (prices.size() == 0) return 0; int maxPrice = prices[prices.size()-1]; int ans = 0; for(int i = prices.size() - 1; i >= 0; i--) { maxPrice = max(maxPrice, prices[i]); ans = max(ans, maxPrice - prices[i]); } return ans; }};
II
Best Time to Buy and Sell Stock II
Total Accepted: 12370 Total Submissions: 34808My SubmissionsSay 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).
int maxProfit(vector<int> &prices) {if (prices.size() <= 1) return 0;int profit = 0;for (int i = 1; i < prices.size(); i++){if (prices[i-1] < prices[i]){profit += prices[i] - prices[i-1];}}return profit; }
Best Time to Buy and Sell Stock III
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).
O(n^2)的算法很容易想到:
找寻一个点j,将原来的price[0..n-1]分割为price[0..j]和price[j..n-1],分别求两段的最大profit。
进行优化:
对于点j+1,求price[0..j+1]的最大profit时,很多工作是重复的,在求price[0..j]的最大profit中已经做过了。
类似于Best Time to Buy and Sell Stock,可以在O(1)的时间从price[0..j]推出price[0..j+1]的最大profit。
但是如何从price[j..n-1]推出price[j+1..n-1]?反过来思考,我们可以用O(1)的时间由price[j+1..n-1]推出price[j..n-1]。
最终算法:
数组l[i]记录了price[0..i]的最大profit,
数组r[i]记录了price[i..n]的最大profit。
已知l[i],求l[i+1]是简单的,同样已知r[i],求r[i-1]也很容易。
最后,我们再用O(n)的时间找出最大的l[i]+r[i],即为题目所求。
class Solution {public: int maxProfit(vector<int> &prices) { // Start typing your C/C++ solution below // DO NOT write int main() function int profit = 0, n = prices.size(); if (n == 0) { return 0; } int l[n], r[n]; memset(l, 0, sizeof(int) * n); memset(r, 0, sizeof(int) * n); int min = prices[0]; for (int i = 1; i < n; i++) { l[i] = prices[i] - min > l[i - 1] ? prices[i] - min : l[i - 1]; min = prices[i] < min ? prices[i] : min; } int max = prices[n - 1]; for (int i = n - 2; i >= 0; i--) { r[i] = max - prices[i] > r[i + 1] ? max - prices[i] : r[i + 1]; max = prices[i] > max ? prices[i] : max; } for (int i = 0; i < n; i++) { profit = l[i] + r[i] > profit ? l[i] + r[i] : profit; } return profit; }};
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