LeetCode-121-Best Time to Buy and Sell Stock(最佳买卖股票的时间)

Q:

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: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.

Analysis:

只需要对数组进行一次遍历，每次记录最小买入价格和最大收益，不断更新即可，时间复杂度为O(n)

Code:

class Solution {    public int maxProfit(int[] prices) {        if (prices.length == 0 || prices.length == 1) {            return 0;        }        int minPrice = prices[0];// 最低买入价格        int maxPofit = 0;// 最高卖出收益        for (int i = 1; i < prices.length; i++) {            // 如果有比当前最低买入价格低的价位，则更新最低买入价格            if (prices[i] < minPrice) {                minPrice = prices[i];            } else {                // 如果有比当前最高收益高的收益，则更新最高收益                if (prices[i] - minPrice > maxPofit) {                    maxPofit = prices[i] - minPrice;                }            }        }        return maxPofit > 0 ? maxPofit : 0;    }}