Easy 121题 Best Time to Buy and Sell Stock

Question:

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.

Everyone said it's a dp problem. Idk. what dynamic programming is.....

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

This way is easy to think of ..However is time limited .......

public class Solution {    public int maxProfit(int[] prices) {        if (prices.length < 2) return 0; //this line is important!!!!!!!        int min=prices[0];        int max_profit=0;        for(int i=0;i<=prices.length-1;i++)        {           min=Math.min(min, prices[i]);           max_profit=Math.max(max_profit,prices[i]-min);        }        return max_profit;    }}

The first line is very important.... in the loop we want to find the min number as the buy prices and check the number behinds if the difference is max~

2016/11/28

public class Solution {    public int maxProfit(int[] prices) {        if(prices==null||prices.length==0)            return 0;        int min=prices[0];        int max=0;        for(int i=1;i<=prices.length-1;i++)        {            if(min>prices[i])                min=prices[i];            max=max>prices[i]-min? max: prices[i]-min;        }        return max;    }}

从左往右遍历，确定找到一个当前最小的值，然后往后用后面的值减，保留最大的值~