第一天

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1. Greedy Algorithm: Best Time to BUY/SELL II

package LeeCodeCrack;

public class DP_BestBuySell2 {

    public static int bestBuySell (int[] price){
        int maxProfit = 0;
        //If the price is not enough, then return 0
        if (price==null||price.length<=1){
            return 0;
        }
        //Make the lastPrice variable as the largest integer exsit, 
        //make sure price[i] will never larger than lastPrice 
        //before lastPrice get the value from price[i]
        int lastPrice=Integer.MAX_VALUE;
        
        //Record every single profit between price[i] and last price
        for (int i=0;i<=price.length-1;i++){
            if(price[i]>lastPrice){
            maxProfit+=price[i]-lastPrice;
            
            }
            //Given the previous value to lastprice
            lastPrice=price[i];
        }
            
        
        return maxProfit;
    }
    
    public static void main (String[] args){
        int[] array={1,2,3,4,5,6};
        System.out.println(bestBuySell(array));
    }

}

2. (5) Stack and Queue: Simplify Path

3. Question not from Leetcode Split 

This should be fairly simple, but I‘m just stuck. Say you have the path /a/b/c/. I‘d like to convert that into an array containing:

• /
• /a/
• /a/b/
• /a/b/c/

•     
  If you need something primitive. Try split and append.
  public class StackOverflow {
  
      public static void main(String args[]) {
  
          String[] folders = "/a/b/c/".split("/");
          String[] paths = new String[folders.length];
          String path = "";
  
          for (int i = 0; i < folders.length; i++) {
              path +=   folders[i] + "/";
              paths[i] = path;
          }
      }
  }
  
  
  run:
  /
  /a/
  /a/b/
  /a/b/c/
  BUILD SUCCESSFUL (total time: 0 seconds)

   4. Stack and Queue (Simplify Path)

• package LeeCodeCrack;
  import java.util.*;
  
  public class Stack_SimlifyPath {
  //解题要点，其实就是知道Java String的split function，
  //然后..的情况用stack，遇到..就把之间压进去的pop出来。最后stack里留下的是反过来的，
  //所以要用另一个stack反回来。边角case就是”//”的情况，空String不压入stack。”.”不压入stack, 
  //遇到”..”只有stack不为空的时候才能pop，否则忽略。没有太多fancy算法。
      public static String simplifyPath (String path){
          String[] paths=path.split("/");
          Stack<String> stack= new Stack<String>();
          for (String s: paths){
              if(s.equals("..")){
                  if(!stack.isEmpty()){
                      stack.pop();
                  }
                  //注意length()是Method
              }else if (s.length()>0 && !s.equals(".")){
                  stack.push(s);
              }
          }
          //重组String
          StringBuilder sb =new StringBuilder();
          for(String s: stack){
              sb.append("/"+s);
          }
          if(sb.length()==0){
              sb.append("/");
          }
          return sb.toString();
          
      }
      
      public static void main (String[] args){
          System.out.println(simplifyPath("/a/b/../../c/"));
      }
  
  }

   

第一天

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原文地址：http://www.cnblogs.com/ChrisY/p/5436029.html