Problem:
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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.
思路：
这题实际上就是求数组后面的元素与前面的元素最大差值。常规的思路是2个for循环遍历一遍，然后求出差值的最大值。但这么做算法复杂度是(O(n^2))。
这题可以借助求最大子串和的算法——Kadane's Algorithm求解。
Solution:
int maxProfit(vector<int>& prices) {
int maxCur = 0, maxSoFar = 0;
for (int i = 1; i < prices.size(); i++) {
maxCur = max(0, maxCur + prices[i]-prices[i-1]);
maxSoFar = max(maxCur, maxSoFar);
}
return maxSoFar;
}
性能：
Runtime: 8 ms Memory Usage: 9.6 MB
