LeetCode Two Sum III

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);
 */