【leetcode系列】121. 买卖股票的最佳时机

给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。

如果你最多只允许完成一笔交易（即买入和卖出一支股票），设计一个算法来计算你所能获取的最大利润。

注意你不能在买入股票前卖出股票。

示例 1:

输入: [7,1,5,3,6,4]
输出: 5
解释: 在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。
     注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格。
示例 2:

输入: [7,6,4,3,1]
输出: 0
解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/*
 * @lc app=leetcode id=121 lang=javascript
 *
 * [121] Best Time to Buy and Sell Stock
 *
 * https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
 *
 * algorithms
 * Easy (46.34%)
 * Total Accepted:    480.5K
 * Total Submissions: 1M
 * Testcase Example:  '[7,1,5,3,6,4]'
 *
 * Say you have an array for which the i^th 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.
 *
 *
 */
/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    let min = prices[0];
    let profit = 0;
    // 7 1 5 3 6 4
    for(let i = 1; i < prices.length; i++) {
        if (prices[i] > prices[i -1]) {
            profit = Math.max(profit, prices[i] - min);
        } else {
            min = Math.min(min, prices[i]);;
        }
    }

    return profit;
};

# 应用Kadane's algorithms
class Solution:
    def maxProfit(self, prices: 'List[int]') -> int:
        """
        step by step
        """
        # error case
        if len(prices) < 1:
            return 0

        # caluate the daily gains, break into a subarray problem
        gains = [prices[i]-prices[i-1] for i in range(1, len(prices))]

        loc_max = global_max = 0 #not gains[0] in case of negative
        for i in range(len(gains)):
            loc_max = max(loc_max + gains[i], gains[i])
            if loc_max > global_max:
                global_max = loc_max
"""
Runtime: 48 ms, faster than 34.50% of Python3 online submissions for Best Time to Buy and Sell Stock.
Memory Usage: 14.1 MB, less than 10.26% of Python3 online submissions for Best Time to Buy and Sell Stock.
"""

/**
 * 系统上C++的测试用例中的输入有[]，因此需要加一个判断
 */
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.empty()) return 0;
        auto min = prices[0];
        auto profit = 0;
        for (auto i = 1; i < prices.size(); ++i) {
            if (prices[i] > prices[i -1]) {
                profit = max(profit, prices[i] - min);
            } else {
                min = std::min(min, prices[i]);;
            }
        }
        return profit;
    }
};