Leetcode 121. Best Time to Buy and Sell Stock – 

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

Input: prices = [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.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

題目解析

給予一個array prices ，每個元素指ith 天的股票價格。算出哪天買哪天賣可以最高獲利

解題步驟

1. 暴力雙迴圈

var maxProfit = function(prices) {
    
    let maxProfit = 0;
    
    for(let i = 0 ; i < prices.length;i++){
        for(let j = i + 1 ; j < prices.length; j++){
            let profit = prices[j] - prices[i]
            if(profit > maxProfit){
                maxProfit = profit;
            }
        }
    }
    
    return maxProfit
};

時間複雜度 O(n^2)，但因為1 <= prices.length <= 105，使用此方法會導致Time Limit Exceeded

2. 根據題目特性，價格要買最低，賣價要賣最高

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    let buy = 10000;
    let max = 0;
    for(let i = 0 ; i < prices.length ; i++){
        if(prices[i] < buy) buy = prices[i]
        if(prices[i] - buy > max) max =  prices[i] - buy
    }
    return max
    
};