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 (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.

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int dif=0;
        int lowpoint=0;
        for(int i=1;i<prices.size();i++){
            int val=prices[i]-prices[lowpoint];
            if(val>dif)
                dif=val;
            else if(val<0)
                lowpoint=i;
        }
        return dif;
    }
};

這道題表面上看似乎複雜度非常高，要去判斷n*n種情況?其實不然，仔細分析 一下，可以得出這樣一個簡單的思路:只需考慮每一次新遍歷到的價格減去最低價，它是否高於之前的價格差（就是利潤），如果高於，那麼就更新利潤。如果新的價格減最低價小於0，那麼就將新價格作爲最低價即可。