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.
題目大意：
在一個數組中，用後面的元素減去前面的元素得到最大值，返回這個最大值。
思路：
可以使用雙循環來出來，但是效率太低。沒有通過。
採用記錄當前之前的最小值，用當前值減去之前最小的值獲得一個臨時最大值，遍歷整個數組，找到最大值。
代碼如下：
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size() <= 1)
return 0;
int max = 0;
int curMin = prices[0];
for(int i = 1;i<prices.size();i++)
{
if(prices[i] - curMin > max)
max = prices[i] - curMin;
if(prices[i] < curMin)
curMin = prices[i];
}
return max;
}
};2016-08-12 08:43:51