• 281. Zigzag Iterator z字型遍历


    [抄题]:

    Given two 1d vectors, implement an iterator to return their elements alternately.

    Example:

    Input:
    v1 = [1,2]
    v2 = [3,4,5,6] 
    
    Output: [1,3,2,4,5,6]
    
    Explanation: By calling next repeatedly until hasNext returns false, 
                 the order of elements returned by next should be: [1,3,2,4,5,6].

    Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?

    Clarification for the follow up question:
    The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example:

    Input:
    [1,2,3]
    [4,5,6,7]
    [8,9]
    
    Output: [1,4,8,2,5,9,3,6,7].

     [暴力解法]:

    时间分析:

    空间分析:

     [优化后]:

    时间分析:

    空间分析:

    [奇葩输出条件]:

    [奇葩corner case]:

    [思维问题]:

    不用写ifv1   if v2.hasnext return true; 直接用temp交换就行了

    [英文数据结构或算法,为什么不用别的数据结构或算法]:

    不知道初始化要写变量 private Iterator<Integer> i, j, tmp;

    [一句话思路]:

    [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

    [画图]:

    [一刷]:

    1. 交换的条件是:i.hasNext(), 别人有下一个才能和你换

    [二刷]:

    [三刷]:

    [四刷]:

    [五刷]:

      [五分钟肉眼debug的结果]:

    [总结]:

    [复杂度]:Time complexity: O(n) Space complexity: O(n)

    [算法思想:迭代/递归/分治/贪心]:

    [关键模板化代码]:

    [其他解法]:

    [Follow Up]:

    [LC给出的题目变变变]:

     [代码风格] :

     [是否头一次写此类driver funcion的代码] :

    实现类的题目不能设为public class,否则要改文件名。

     [潜台词] :

     

    // package whatever; // don't place package name!
    
    
    import java.io.*;
    import java.util.*;
    import java.lang.*;
    /*
    Input:
    v1 = [1,2]
    v2 = [3,4,5,6] 
    
    Output: [1,3,2,4,5,6]
    */
    class driverFuction {
      public static void main (String[] args) {
        List<Integer> v1 = Arrays.asList(1, 2);
        List<Integer> v2 = Arrays.asList(3, 4, 5, 6);
        ZigzagIterator answer = new ZigzagIterator(v1, v2);
        System.out.println(answer.next());
        System.out.println(answer.next());
        System.out.println(answer.next());
        System.out.println(answer.next());
        System.out.println(answer.next());
      }
    }
    
    class ZigzagIterator {
        Iterator<Integer> i, j, temp;
      
        public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
            i = v1.iterator();
            j = v2.iterator();
        }
    
        public int next() {
            if (i.hasNext()) {temp = j; j = i; i = temp;}
            return j.next();
        }
    
        public boolean hasNext() {
            if (i.hasNext() || j.hasNext()) return true;
          return false;
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/immiao0319/p/9463828.html
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