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  • LeetCode_122. Best Time to Buy and Sell Stock II

    122. Best Time to Buy and Sell Stock II

    Easy

    Say you have an array for which the ith element is the price of a given stock on day i.

    Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

    Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

    Example 1:

    Input: [7,1,5,3,6,4]
    Output: 7
    Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
                 Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
    

    Example 2:

    Input: [1,2,3,4,5]
    Output: 4
    Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
                 Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
                 engaging multiple transactions at the same time. You must sell before buying again.
    

    Example 3:

    Input: [7,6,4,3,1]
    Output: 0
    Explanation: In this case, no transaction is done, i.e. max profit = 0.
    package leetcode.easy;
    
    public class BestTimeToBuyAndSellStockII {
    	@org.junit.Test
    	public void test() {
    		int[] prices1 = { 7, 1, 5, 3, 6, 4 };
    		int[] prices2 = { 1, 2, 3, 4, 5 };
    		int[] prices3 = { 7, 6, 4, 3, 1 };
    		System.out.println(maxProfit1(prices1));
    		System.out.println(maxProfit1(prices2));
    		System.out.println(maxProfit1(prices3));
    		System.out.println(maxProfit2(prices1));
    		System.out.println(maxProfit2(prices2));
    		System.out.println(maxProfit2(prices3));
    		System.out.println(maxProfit3(prices1));
    		System.out.println(maxProfit3(prices2));
    		System.out.println(maxProfit3(prices3));
    	}
    
    	public int maxProfit1(int[] prices) {
    		return calculate(prices, 0);
    	}
    
    	public int calculate(int prices[], int s) {
    		if (s >= prices.length) {
    			return 0;
    		}
    		int max = 0;
    		for (int start = s; start < prices.length; start++) {
    			int maxprofit = 0;
    			for (int i = start + 1; i < prices.length; i++) {
    				if (prices[start] < prices[i]) {
    					int profit = calculate(prices, i + 1) + prices[i] - prices[start];
    					if (profit > maxprofit) {
    						maxprofit = profit;
    					}
    				}
    			}
    			if (maxprofit > max) {
    				max = maxprofit;
    			}
    		}
    		return max;
    	}
    
    	public int maxProfit2(int[] prices) {
    		if (null == prices || 0 == prices.length) {
    			return 0;
    		}
    		int i = 0;
    		int valley = prices[0];
    		int peak = prices[0];
    		int maxprofit = 0;
    		while (i < prices.length - 1) {
    			while (i < prices.length - 1 && prices[i] >= prices[i + 1]) {
    				i++;
    			}
    			valley = prices[i];
    			while (i < prices.length - 1 && prices[i] <= prices[i + 1]) {
    				i++;
    			}
    			peak = prices[i];
    			maxprofit += peak - valley;
    		}
    		return maxprofit;
    	}
    
    	public int maxProfit3(int[] prices) {
    		int maxprofit = 0;
    		for (int i = 1; i < prices.length; i++) {
    			if (prices[i] > prices[i - 1]) {
    				maxprofit += prices[i] - prices[i - 1];
    			}
    		}
    		return maxprofit;
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/denggelin/p/11629181.html
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