描述
Say you have an array for which the i-th 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
解法一:使用贪心策略,分别找到价格最低和最高的一天,低进高出,注意最低的一天要在最高的一天之前。时间复杂度为O(n), 空间复杂度为O(1).
代码如下
class Solution {
public:
int maxProfit(vector<int> &prices) {
if(prices.size() < 2) return 0;
int profit = 0;
int cur_min = prices[0];
for (int i = 1; i < prices.size(); i++) {
profit = max(profit, prices[i] - cur_min);
cur_min = min(cur_min, prices[i]);
}
return profit;
}
}
解法二:把原始价格序列变成差分序列,本题也可以看做是最大m子段和,m=1。