【LeetCode】1. Two Sum

Difficulty: Easy

 More:【目录】LeetCode Java实现

Description

https://leetcode.com/problems/two-sum/

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactlyone solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Intuition

1.Brute Force ：Time complexity O(n^2); Space complexity O(1);   ×

2.Making the use of HashMap to reduce time consumption, (If the complement exists in the array, we need to look up its index. A hash table is the best way to looke up). <nums, index>  ==> <key, value>

Solution

    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();       
        for(int i=0; i<nums.length; i++){
            if(map.containsKey(target-nums[i]))
                return new int[]{i,map.get(target-nums[i])};
            map.put(nums[i],i);
        }
        return null;
    }

　　

　　

Complexity

Time complexity : O(n)

Space complexity : O(n)

 

What I've learned

1. How to use HashMap:
　　* map.put(key, value)   ……not "add(K,V)"

　　* map.containsKey(key)

　　* map.containsValue(value)

　　* map.get(key)

　　* map.remove(key)

　　* new HashMap<K,V>();  ……two generics ,not one

2. When we need to find something quickly, it's a good way to use HashMap.

3. Be careful when input is [3,2,4] and 6, the output should be [1,2] instead of [0,0] .

 More:【目录】LeetCode Java实现