【LeedCode】121.Best Time to Buy and Sell Stock

Description:

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: 5

max. 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: 0

In this case, no transaction is done, i.e. max profit = 0.

题目分析：简单的买卖股票，遇到最小的就保存表示买进，计算后面与它的差值取最大差值即为收益。若后面的数值都比它小，则收益为零。

Solutions:

class Solution {

public:

    int maxProfit(vector<int>& prices) {

if(prices==null||prices.length<=1){ 

return 0; 

  }

  int res=0,temp,inv;

temp=prices[0];

for(int i=1;i<prices.length;i++){

inv=prices[i];

    if(temp>inv){

    temp=inv;

}

else{

inv=inv-temp;

res=res>inv?res:inv;

}

}

return res;

    }

};