"Leetcode" best time to Buy and Sell Stock

Test instructions

Say you had an array for which the ith element was the price of a given stock on day I.

If you were-permitted-to-complete at most one transaction (ie, buy one and sell one share of the stock), design an AL Gorithm to find the maximum profit.

Ideas:

can only buy and sell once, so to find a maximum difference of two days, but pay attention to the low that day is in front. So you can maintain a minimum and a maximum difference, and then go backward, if it is smaller than the minimum value is updated, if the difference between the minimum value is greater than the previous maximum difference, the maximum difference is updated.

Note that if you maintain the maximum and minimum values, and finally return a maximum-minimum number, it will be wrong oh!

Code: C + +:

Class Solution {public:    int maxprofit (vector<int> &prices) {int n = prices.size (); if (n <= 1) return 0;int Min_value = prices[0];int ans = 0;for (Vector<int>::iterator it = Prices.begin (); It! = Prices.end (); it++) {/* Update minimum * /if (*it < min_value) {min_value = *it;} /* Update max Difference */if (*it-min_value > Ans) {ans = *it-min_value;}} return ans;            };

Python:

Class solution:    # @param prices, a list of integer    # @return An integer    def maxprofit (self, prices):        if Le N (Prices) < 2:            return 0        min = prices[0]        Profit = 0 for        x in prices:            if x < min:                min = x
   if x-min > Profit:                Profit = x-min        return profit

"Leetcode" best time to Buy and Sell Stock