LeetCode Two Sum III - Data structure design

原题链接在这里：https://leetcode.com/problems/two-sum-iii-data-structure-design/

题目：

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

题解：

与Two Sum类似.

设计题目考虑trade off. 如果add 很多, find很少可以采用map方法.

简历HashMap保存key 和 key的frequency. find value时把HashMap的每一个key当成num1, 来查找HashMap是否有另一个value-num1的key.

num1 = value - num1时，检查num1的frequency是否大于1.

Time Complexity: add O(1). find O(n). Space: O(n).

AC Java:

 class TwoSum {

     HashMap<Integer, Integer> hm;

     /** Initialize your data structure here. */
     public TwoSum() {
         hm = new HashMap<Integer, Integer>();
     }

     /** Add the number to an internal data structure.. */
     public void add(int number) {
         hm.put(number, hm.getOrDefault(number, 0) + 1);
     }

     /** Find if there exists any pair of numbers which sum is equal to the value. */
     public boolean find(int value) {
         for(Map.Entry<Integer, Integer> entry : hm.entrySet()){
             int num1 = entry.getKey();
             int num2 = value-num1;
             if(hm.containsKey(num2) && (hm.get(num2)>1 || num1!=num2)){
                 return true;
             }
         }
         return false;
     }
 }

 /**
  * Your TwoSum object will be instantiated and called as such:
  * TwoSum obj = new TwoSum();
  * obj.add(number);
  * boolean param_2 = obj.find(value);
  */

考虑到trade off, 上面用HashMap时add用O(1), find用O(n).

用两个Set分别存现有nums 和现有sums可以treade off.

这种设计适合少量add, 多量find操作.

Time Complexity: add O(n). find O(1). Space: O(n).

AC Java:

 public class TwoSum {
     HashSet<Integer> nums;
     HashSet<Integer> sums;

     /** Initialize your data structure here. */
     public TwoSum() {
         nums = new HashSet<Integer>();
         sums = new HashSet<Integer>();
     }

     /** Add the number to an internal data structure.. */
     public void add(int number) {
         Iterator<Integer> it = nums.iterator();
         while(it.hasNext()){
             sums.add(it.next()+number);
         }
         nums.add(number);
     }

     /** Find if there exists any pair of numbers which sum is equal to the value. */
     public boolean find(int value) {
         return sums.contains(value);
     }
 }

 /**
  * Your TwoSum object will be instantiated and called as such:
  * TwoSum obj = new TwoSum();
  * obj.add(number);
  * boolean param_2 = obj.find(value);
  */