best-time-to-buy-and-sell-stock

客官°小女子只卖身不卖艺 2021-11-09 08:34 164阅读 0赞

/\*\*  
\*  
\* @author gentleKay  
\* 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 (ie, buy one and sell one share of the stock),  
\* design an algorithm to find the maximum profit.  
\*  
\* 假设您有一个数组，其中第i个元素是第一天给定股票的价格。  
\* 如果你只被允许完成最多一笔交易（即买卖一份股票），  
\* 设计一个算法来找到最大利润。  
\*/

#### 第一种方法： ####

/**
     * 
     * @author gentleKay
     * 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 (ie, buy one and sell one share of the stock), 
     * design an algorithm to find the maximum profit.
     * 
     * 假设您有一个数组，其中第i个元素是第一天给定股票的价格。
     * 如果你只被允许完成最多一笔交易（即买卖一份股票），
     * 设计一个算法来找到最大利润。
     */
    
    
    public class Main11 {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    		int[] prices = {1};
    		System.out.println(Main11.maxProfit(prices));
    	}
    	
    	public static int maxProfit(int[] prices) {
            int min = Integer.MAX_VALUE;
            int res = 0;
            for (int i=0;i<prices.length;i++) {
            	min = Math.min(min, prices[i]);
            	res = Math.max(res, prices[i] - min);
            }
            return res;
        }
    
    }

#### 第二种方法： ####

import java.util.ArrayList;
    import java.util.Collections;
    
    /**
     * 
     * @author gentleKay
     * 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 (ie, buy one and sell one share of the stock), 
     * design an algorithm to find the maximum profit.
     * 
     * 假设您有一个数组，其中第i个元素是第一天给定股票的价格。
     * 如果你只被允许完成最多一笔交易（即买卖一份股票），
     * 设计一个算法来找到最大利润。
     */
    
    
    public class Main11 {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    		int[] prices = {1};
    		System.out.println(Main11.maxProfit(prices));
    	}
    	
    	public static int maxProfit(int[] prices) {
    		
    		if (prices.length < 1) {
    			return 0;
    		}
    		ArrayList<Integer> array = new ArrayList<>();
    		for (int i=0;i<prices.length;i++) {
    			for (int j=i+1;j<prices.length;j++) {
    				if (prices[j] - prices[i] < 0) {
    					continue;
    				}
    				array.add(prices[j] - prices[i]);
    			}
    		}
    		Collections.sort(array);
    		if (array.isEmpty()) {
    			return 0;
    		}
    		return array.get(array.size()-1);
        }
    
    }

转载于:https://www.cnblogs.com/strive-19970713/p/11244796.html

best-time-to-buy-and-sell-stock

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