LeetCode 121. Best Time to Buy and Sell Stock
2016-05-31 22:05
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Best Time to Buy and Sell Stock
Des
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.
分析
看到题,自己思路很清晰,这个存在一个时间序列问题在里边就是,先买后卖,卖的智能位于买的后边,网上说是动态规划,我不是很懂,就先按自己思路写了一个。先找一个目前最小的,从当前之后找一个最大的,复杂度N**3了,太怕。class Solution { public: int max1(vector<int>& prices,int i){ int maxp=-1000; for(;i<prices.size();i++){ if(prices[i]>maxp) maxp=prices[i]; } return maxp; } int min1(vector<int>& prices,int i){ int minp=1000; for(int j=0;j<=i;j++){ if(prices[j]<minp) minp=prices[i]; } return minp; } int maxProfit(vector<int>& prices) { int maxp=-1000; for(int i=0;i<prices.size();i++){ if((max1(prices,i) - min1(prices,i)) > maxp) maxp=(max1(prices,i)-min1(prices,i)); } return maxp; } };
自己写的也能work,这也算不赖了,但是有些情况会超时,并且题是easy的不应该有这么多代码,可是自己真没什么思路了。
别人处理的方法
首先从后往前找,边找出最大的卖出值,边求最大的卖出价,这个题的code真的觉得很精妙。Code
class Solution { public: int maxProfit(vector<int>& prices) { int ans=-1,res=0; for(int i=prices.size()-1;i>=0;i--){ ans=max(ans,prices[i]); res=max(res,ans-prices[i]); } return res; } };
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