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.
Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5
max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.
分析：
動(dòng)態(tài)規(guī)劃法弟头。從前向后遍歷數(shù)組涉茧，記錄當(dāng)前出現(xiàn)過的最低價(jià)格，作為買入價(jià)格伴栓，并計(jì)算以當(dāng)天價(jià)格出售的收益伦连，作為可能的最大收益惑淳，整個(gè)遍歷過程中，出現(xiàn)過的最大收益就是所求饺窿。時(shí)間復(fù)雜度O(n)，空間復(fù)雜度O(1)肚医。
具體代碼如下：
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.empty()) return 0;
int minPrice = prices[0];
int maxProfit = 0;
for(int i = 1; i < prices.size();++i)
{
maxProfit = max(maxProfit,prices[i] - minPrice);
minPrice = min(minPrice,prices[i]);
}
return maxProfit;
}
};
