Best Time to Buy and Sell Stock III
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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.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
现在条件变成了最多两次,一个直接的思路是枚举切分点,然后分别计算前后只能买一次的最大值,相加;当然也可能整个只买一次。
直观看复杂度是o(n2)的,试了下大数据果然是过不了~
其实遇到这种最多两次的题,要敏感,最多两次跟最多K次还不一样,最多两次的题通常都可以转化为一次的情况,即前向扫描一遍,保留profit[i],再后向扫描一遍,扫描到i的时候,就是这次扫描的最大值加前向扫描的那个profit[i]之和。
代码:
class Solution {public: int maxProfit(vector<int>& price){int n=price.size();if (n<=1)return 0;vector<int> profit(n,0);int low=price[0],i;for(i=1;i<n;i++){if (price[i]>low){profit[i]=max(profit[i-1],price[i]-low);}else{profit[i]=profit[i-1];low=price[i];}}int high=price[n-1];int ans=0;for(i=n-1;i>=0;i--){if (high>price[i]){ans=max(ans,profit[i]+high-price[i]);}elsehigh=price[i];}return ans;}};
人家一样的思路为什么16ms过大合集,我的要60ms啊!!!!!
是因为使用了max函数吗? 哪位同学知道
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