Best Time to Buy and Sell Stock
Say you have an array for which theithelement is the price of a given stock on dayi.
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.
题意：做一次买入卖出的最大收益。
思路：一次线性扫描解决。
public class Solution {public int maxProfit(int[] prices) {if (prices.length == 0) return 0;int low = prices[0];int ans = 0;for (int i = 1; i < prices.length; i++) {if (prices[i] < low) low = prices[i];else if (ans < prices[i] – low) ans = prices[i] – low;}return ans;}}
Best Time to Buy and Sell Stock II
Say you have an array for which theithelement is the price of a given stock on dayi.
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).
题意：可以不断买入卖出的最大收益。
思路：按理说找出每段上升序列的头尾差，但是可以发现其实头尾差就是每相邻的差(保证后一位较大)。
public class Solution {public int maxProfit(int[] prices) {if (prices.length == 0) return 0;int ans = 0;for (int i = 0; i < prices.length – 1; i++) {if (prices[i+1] > prices[i])ans += prices[i+1] – prices[i];}return ans;}}
Best Time to Buy and Sell Stock III
Say you have an array for which theithelement is the price of a given stock on dayi.
Design an algorithm to find the maximum profit. You may complete at mosttwotransactions.
Note:You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
题意：只能买入卖出两次的最大收益。
思路：注意卖出和买入可以同一时刻，两次就意味着把序列切成了两部分，那么一次从前遍历执行第一题的操作，再从后来一次，求和的最大值。
public class Solution {public int maxProfit(int[] prices) {if (prices.length == 0) return 0;final int len = prices.length;int low = prices[0], Max = 0;int[] profit = new int[len];profit[0] = 0;for (int i = 1; i < len; i++) {low = Math.min(low, prices[i]);if (Max < prices[i] – low) Max = prices[i] – low;profit[i] = Max;}int tmpMax = prices[len-1];int ans = profit[len-1];Max = 0;for (int i = len-2; i >= 0; i–) {tmpMax = Math.max(tmpMax, prices[i]);if (Max < tmpMax – prices[i])Max = tmpMax – prices[i];if (ans < profit[i] + Max)ans = profit[i] + Max;}return ans;}}
版权声明：本文为博主原创文章，未经博主允许不得转载。
，怪天怪地，我都不会怪你，你有选择幸福的权利…
