Array - 121. Best Time to Buy and Sell Stock

ppxai

发布于 2020-09-23 17:07:40

2590

发布于 2020-09-23 17:07:40

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.

思路：

这一题使用dp来做，实质也就是找到当前价之前的最低价，来判断盈利，也可以转换为求最大子数组和（53题）来做，也可以用贪心，直接找出最低价与当前盈利。

代码：

java：

class Solution {

    // 1. time:O(n) space:O(0)
    /*public int maxProfit(int[] prices) {
        int len = prices.length;
        if (len < 1) return 0;
        
        int[] minPrices = new int[len];
        int[] maxProfit = new int[len];
        minPrices[0] = prices[0];
        maxProfit[0] = 0;
        
        for (int i = 1; i < len; i++) {
            minPrices[i] = Math.min(minPrices[i-1], prices[i]);
            maxProfit[i] = Math.max(maxProfit[i-1], prices[i] - minPrices[i-1]);
        }
        
        return maxProfit[len-1];
    }*/
    
    // 2. time: O(n) space:O(1)
    public int maxProfit(int[] prices) {
        int len = prices.length;
        if (len < 1) return 0;
        
        int profit = 0;
        int buyPrice = prices[0];
        
        for (int i = 1; i < len; i++) {
            buyPrice = Math.min(prices[i], buyPrice);
            profit = Math.max(prices[i] - buyPrice, profit);
        } 
            
        return profit;
    }
}

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2019年06月15日，如有侵权请联系 cloudcommunity@tencent.com 删除

case