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;
}
}

LeetCode_122. Best Time to Buy and Sell Stock II的更多相关文章

  1. [LintCode] Best Time to Buy and Sell Stock II 买股票的最佳时间之二

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

  2. LEETCODE —— Best Time to Buy and Sell Stock II [贪心算法]

    Best Time to Buy and Sell Stock II Say you have an array for which the ith element is the price of a ...

  3. 27. Best Time to Buy and Sell Stock && Best Time to Buy and Sell Stock II && Best Time to Buy and Sell Stock III

    Best Time to Buy and Sell Stock (onlineJudge: https://oj.leetcode.com/problems/best-time-to-buy-and- ...

  4. Leetcode-122 Best Time to Buy and Sell Stock II

    #122  Best Time to Buy and Sell Stock II Say you have an array for which the ith element is the pric ...

  5. 【leetcode】Best Time to Buy and Sell Stock II

    Best Time to Buy and Sell Stock II Say you have an array for which the ith element is the price of a ...

  6. 31. leetcode 122. Best Time to Buy and Sell Stock II

    122. Best Time to Buy and Sell Stock II Say you have an array for which the ith element is the price ...

  7. LeetCode: Best Time to Buy and Sell Stock II 解题报告

    Best Time to Buy and Sell Stock IIQuestion SolutionSay you have an array for which the ith element i ...

  8. Algorithm - 贪心算法使用场景 ( LEETCODE —— Best Time to Buy and Sell Stock II)

    先看一道leetcode题: Best Time to Buy and Sell Stock II Say you have an array for which the ith element is ...

  9. leetcode 121. Best Time to Buy and Sell Stock 、122.Best Time to Buy and Sell Stock II 、309. Best Time to Buy and Sell Stock with Cooldown

    121. Best Time to Buy and Sell Stock 题目的要求是只买卖一次,买的价格越低,卖的价格越高,肯定收益就越大 遍历整个数组,维护一个当前位置之前最低的买入价格,然后每次 ...

随机推荐

  1. phpstorm快捷键使用

  2. redis 与 序列化

    概念 序列化:把对象转化为可传输的字节序列过程称为序列化. 反序列化:把字节序列还原为对象的过程称为反序列化. 为什么需要序列化 序列化最终的目的是为了对象可以跨平台存储,和进行网络传输.而我们进行跨 ...

  3. Base 编解码(转)

    private static final char[] legalChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0 ...

  4. [Cypress] install, configure, and script Cypress for JavaScript web applications -- part4

    Load Data from Test Fixtures in Cypress When creating integration tests with Cypress, we’ll often wa ...

  5. SpringMVC拦截器及多拦截器时的执行顺序

    本文链接:https://blog.csdn.net/itcats_cn/article/details/80371639拦截器的配置步骤 springmvc.xml中配置多个拦截器配置自定义拦截器并 ...

  6. GridView修改含有DropDownList控件列的宽度

    GridView进入Edit模式,编辑列动态绑定DropDown List方便客户选择,但当里面的Item过长,不免令界面不美观 正确做法: <asp:TemplateField HeaderT ...

  7. 齿轮 HYSBZ - 4602 (DFS实现)

    齿轮 HYSBZ - 4602 题意:很好理解就不啰嗦了. 致谢:感谢队友小明. 题解:嗯,一开始想到的是并查集,后来,就先看了另一道题,xj写dfs和暴力,就卡死了.于是来补这题了,前向星建图 题解 ...

  8. C++标准库分析总结(三)——<迭代器设计原则>

    本节主要总结迭代器的设计原则,以及iterstor traits的设计作用 1.迭代器遵循的原则 迭代器是算法和容器的桥梁,它是类模板的设计,迭代器必须有能力回答算法提出的问题才能去搭配该算法的使用 ...

  9. css3 perspective与translateZ变换

    css3中的坐标系,rotateX就是绕着x轴旋转,rotateY就是绕着Y轴旋转,rotateZ就是绕着z轴旋转(也就是xy平面的旋转). perspective属性用来设置视点,在css3的模型中 ...

  10. 2019-06-03 校内python模拟题解(所有非原题)

    一起来女装吧 本题改编自USACO(USA Computing Olympiad) 1.1节的第一题 (感谢lsy同学对本题题面的贡献) 直接计算就好了 chr:将ASCII码转成字符 ord:字符对 ...