[LeetCode] 121. Best Time to Buy and Sell Stock ❤

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.

______________________________________________ 题目

写了两次，第一版两次循环有一个样例time limited了；第二次改成一次循环
第二版思路：每天的价格既可以是买入价也可以是卖出价，所以每次循环，如果可以就更新最低买入价；否则，就更新最大收入（如果可以）

int maxProfit(vector<int>& prices) {
if (prices.empty()) return 0;
int max = 0;
for (int i = 0; i < prices.size() - 1; ++i) {
for (int j = i + 1; j < prices.size(); ++j) {
if (prices[i] < prices[j]) {
if (prices[j] - prices[i] > max)
max = prices[j] - prices[i];
}
}
}
return max;
}

int maxProfit(vector<int>& prices) {
if (prices.empty() || prices.size() == 1) return 0;
int max = 0;
int min = prices[0];
for (int i = 1; i < prices.size(); ++i) {
if (prices[i] < min) {
min = prices[i];
}
else {
if (prices[i] - min > max)
max = prices[i] - min;
}
}
return max;
}