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.
用一个数组A[n-1]来存储当天与前一天股票的差价，从而将该题转化为最大子序列和的问题。仅仅只是没有负值。最小为0罢了。
A[i]表示第i+1天股票价格与第i天的股票价格的差值。
public class Solution {public int maxsum(int [] num){int sum=0;int max=Integer.MIN_VALUE;for(int i=0;i<num.length;i++){sum+=num[i];if(max<sum) max=sum;if(sum<0) sum=0;}return max>0?max:0;}public int maxProfit(int[] prices) {if(prices.length==0||prices.length==1) return 0;int [] sub=new int [prices.length-1];for(int i=0;i<sub.length;i++){sub[i]=prices[i+1]-prices[i];}return maxsum(sub);}}
