问题描述:
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.
问题分析:利用动态规划的思想,从前向后遍历数组,记录当前出现过的最低的价格,做为买入时候的价格,那么直接遍历整个数组,看哪一天卖出即可,选择哪一天利润最大的卖出。
代码:
class Solution {
public:
int max(int a,int b){
return a>b?a:b;
}
int min(int a,int b){
return a<b?a:b;
}
int maxProfit(vector<int>& prices) {
if(prices.size()<2) return 0;
int maxProfit=0;
int curMin=prices[0];
for (int i=1;i<prices.size();i++){
curMin=min(curMin,prices[i]);
maxProfit=max(maxProfit,prices[i]-curMin);
}
return maxProfit;
}
};