Best Time to Buy and Sell Stock 系列问题

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.

Example 1:

Input: [7, 1, 5, 3, 6, 4]Output: 5max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]Output: 0In this case, no transaction is done, i.e. max profit = 0.

这是一个模范做法= =我自己做的超时了。。。。。。。。

这是买股票的问题 首先要在前面选择一个最低价买入 然后要在买入之后选择最大利润的时候卖出

股票是一个系列问题 日后继续更

股票系列问题

public class Solution {    public int maxProfit(int[] prices) {        if (prices.length < 2) return 0;                int maxProfit = 0;        int curMin = prices[0];                for (int i = 1; i < prices.length; i++) {//遍历一次数组 寻找            curMin = Math.min(curMin, prices[i]);寻找目前遍历过的最低价            maxProfit = Math.max(maxProfit, prices[i] - curMin);        }                return maxProfit;    }}

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 (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).

可以买卖多次 使用贪心法的思维 只要有利润可以赚就卖出

public class Solution {    public int maxProfit(int[] prices) {             int maxprofit=0;        for(int i=1;i<prices.length;i++){            if(prices[i]>prices[i-1]){                maxprofit+=prices[i]-prices[i-1];            }        }        return maxprofit;    }}

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).

限定买卖次数为两次的情况   将分别遍历在不同位置分为两半的买卖利润

选取最大值

public class Solution {    public int maxProfit(int[] prices) {        if (prices.length < 2) return 0;                int n = prices.length;        int[] preProfit = new int[n];        int[] postProfit = new int[n];                int curMin = prices[0];        for (int i = 1; i < n; i++) {            curMin = Math.min(curMin, prices[i]);            preProfit[i] = Math.max(preProfit[i - 1], prices[i] - curMin);        }                int curMax = prices[n - 1];        for (int i = n - 2; i >= 0; i--) {            curMax = Math.max(curMax, prices[i]);            postProfit[i] = Math.max(postProfit[i + 1], curMax - prices[i]);        }                int maxProfit = 0;        for (int i = 0; i < n; i++) {            maxProfit = Math.max(maxProfit, preProfit[i] + postProfit[i]);        }                return  maxProfit;    }}