题意:有一个数组,第i个数据代表的是第i天股票的价格,每天只能先卖出再买进(可以不卖出也可以不买进),求最大收益。

思路:自己去弄几个数组比划比划就知道了,比如[1,2,5,3,6],第一天买进,第二天卖出,再买进,第三天卖出,第四天买进,第五天卖出。

真正计算的就是前一天的价格和当天的价格的差值,[1,3,-2,3],大于0买进,否则不买。

代码:

int maxProfit(vector<int>& prices) {
int n = prices.size();
int a[n-];
for(int i = ;i<n-;i++)
a[i] = prices[i+]-prices[i];
int ans = ;
for(int j = ;j<n-;j++)
{
if(a[j]>)
ans += a[j];
}
return ans;
}

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