• Leetcode: Two Sum III


    Design and implement a TwoSum class. It should support the following operations: add and find.
    
    add - Add the number to an internal data structure.
    find - Find if there exists any pair of numbers which sum is equal to the value.
    
    For example,
    add(1); add(3); add(5);
    find(4) -> true
    find(7) -> false

    The trade off should be considered: In fact, there has to be one operation's time complexity is O(n) and the other is O(1), no matter add or find. So clearly there's trade off when solve this problem, prefer quick find or quick add.

    if there are more find, add can be pre-done

     1 public class TwoSum {
     2         Set<Integer> sum;
     3         Set<Integer> num;
     4         
     5         TwoSum(){
     6             sum = new HashSet<Integer>();
     7             num = new HashSet<Integer>();
     8         }
     9         // Add the number to an internal data structure.
    10         public void add(int number) {
    11             if(num.contains(number)){
    12                 sum.add(number * 2);
    13             }else{
    14                 Iterator<Integer> iter = num.iterator();
    15                 while(iter.hasNext()){
    16                     sum.add(iter.next() + number);
    17                 }
    18                 num.add(number);
    19             }
    20         }
    21     
    22         // Find if there exists any pair of numbers which sum is equal to the value.
    23         public boolean find(int value) {
    24             return sum.contains(value);
    25         }
    26     }

     more add:

     1 public class TwoSum {
     2     HashMap<Integer, Integer> map;
     3     public TwoSum() {
     4         map = new HashMap<Integer, Integer>();
     5     }
     6     
     7     public void add(int x) {
     8         if (map.containsKey(x)) {
     9             map.put(x, map.get(x)+1);
    10         }
    11         else {
    12             map.put(x, 1);
    13         }
    14     }
    15     
    16     public boolean find(int target) {
    17         for (int i : map.keySet()) {
    18             if (map.containsKey(target-i)) {
    19                 if (target - i != i) return true;
    20                 else if (map.get(i) >= 2) return true;
    21             }
    22         }
    23         return false;
    24     }
    25 }

    注意17行的map.KeySet() return是一个Set形式的key的集合,Set是Collection的Subinterface, 所以这种for循环方法对Set也适用。而且即使key理论上是Integer, for循环前面的元素还是可以写成int:for(int i : map.keySet())

  • 相关阅读:
    Ajax 传递json字符串到客户端时报 Internal server error
    Java 判断字符串的存储长度
    5个数组Array方法: indexOf、filter、forEach、map、reduce使用实例
    databales详解(一)
    JQuery总结
    《JavaScript 高级程序设计》总结
    ASP.NET MVC中controller和view相互传值的方式
    Jquery tmpl详解
    @section script{}的使用
    messager(消息窗口)
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/4252598.html
Copyright © 2020-2023  润新知