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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
难度:medium
思路: 动态规划问题。i处能够获得的最大利润i之后的最大值-i处的price。循环通过倒序可得O(n)的解法
代码如下
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()<=1)
return 0;
int maxPrice = prices[prices.size()-1];
int maxprofit=0;
for(int i=prices.size()-1;i>=0;i--){
maxPrice = max(maxPrice,prices[i]);
maxprofit=max(maxPrice-prices[i],maxprofit);
}
return maxprofit;
}
};
本文介绍如何使用动态规划解决买卖股票问题,找到在给定价格数组中完成最多一次交易的最大收益。
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