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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.
Given an example [3,2,3,1,2], return 1
public class Solution { /** * @param prices: Given an integer array * @return: Maximum profit */ public int maxProfit(int[] prices) { // write your code here if(prices.length==0 || prices==null) return 0; int n=prices.length;int max=0; int low=prices[0]; for(int i=1;i<n;i++) { if(prices[i]<low) { low=prices[i]; } else { max=Math.max(max,prices[i]-low); } } return max; } }
Best Time to Buy and Sell Stock II
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).
Given an example [2,1,2,0,1], return 2
class Solution { /** * @param prices: Given an integer array * @return: Maximum profit */ public int maxProfit(int[] prices) { // write your code here if(prices.length==0 || prices==null) return 0; int n=prices.length; int sum=0; for(int i=0;i<n-1;i++) { if(prices[i+1]>prices[i]) { sum=sum+(prices[i+1]-prices[i]); } } return sum; } };
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.
Given an example [4,4,6,1,1,4,2,5], return 6.
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
class Solution { /** * @param prices: Given an integer array * @return: Maximum profit */ public int maxProfit(int[] prices) { // write your code here if(prices.length==0 || prices==null||prices.length==1) return 0; int n=prices.length; int[] opt=new int[n]; int[] optReverse=new int[n]; int ans=0; opt[0]=0; int low=prices[0]; int result=0; for(int i=1;i<n;i++) { if(prices[i]<low) { low=prices[i]; } else { result=Math.max(prices[i]-low,result); } opt[i]=result; } optReverse[n-1]=0; int high=prices[n-1]; int result1=0; for(int i=n-2;i>=0;i--) { if(prices[i]>high) { high=prices[i]; } else { result1=Math.max(high-prices[i],result1); } optReverse[i]=result1; } for(int i=0;i<n;i++) { int max=opt[i]+optReverse[i]; ans=Math.max(max,ans); } return ans; } };
lintcode medium Best Time to Buy and Sell Stock I,II,III
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原文地址：http://www.cnblogs.com/kittyamin/p/5055909.html
