这道题就是提个醒：有个运行时间更少的办法，减少不必要的循环次数。
题目：
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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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 Explanation: In this case, no transaction is done, i.e. max profit = 0.
解决办法如下：
不好的：
package test;
public class LC121Try1
{
//601ms(so long)
public int maxProfit(int[] prices)
{
int max = 0;
int len = prices.length;
if(len<=1){
return max;
}
for(int i=0;i<len-1;i++){
for(int j=i+1;j<len;j++){
if(prices[j]-prices[i]>max){
max=prices[j]-prices[i];
}
}
}
return max;
}
}
好的解决办法：
package test;
public class LC121Try2
{
//1ms 其实关键是有顺序是有关系的，j永远在i后，所以可以减少循环次数
public int maxProfit(int[] prices)
{
int max = 0;
int len = prices.length;
if (len <= 1)
{
return max;
}
int min = prices[0];
for (int i = 1; i < len; i++)
{
if (prices[i] < min)
{
min = prices[i];
}
else
{
if ((prices[i] - min) > max)
{
max = prices[i] - min;
}
}
}
return max;
}
}
哈哈