【LeetCode刷题记录】1.Two Sum解法与Hashmap的应用

Description:

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 exactly one 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].

My solution:

 public int[] twoSum(int[] nums, int target) {
    for(int i = 0; i < nums.length; i++){
        for(int j = 0; j < nums.length; j++){
            if(nums[i] + nums[j] == target && i!=j){
                return new int [] {i, j};
            }
        }
    }
    return null;
}

Runtime：56ms
Your runtime beats 9.43% of java submissions.

显然时间复杂度为O(n^2)，耗时太长，不太合理。

Better Solutions:

以下为其他人提交的时间复杂度为O(n)的算法。

C++

vector<int> twoSum(vector<int> &numbers, int target)
{
    //Key is the number and value is its index in the vector.
    unordered_map<int, int> hash;
    vector<int> result;
    for (int i = 0; i < numbers.size(); i++) {
        int numberToFind = target - numbers[i];

            //if numberToFind is found in map, return them
        if (hash.find(numberToFind) != hash.end()) {
                    //+1 because indices are NOT zero based
            result.push_back(hash[numberToFind] + 1);
            result.push_back(i + 1);            
            return result;
        }

            //number was not found. Put it in the map.
        hash[numbers[i]] = i;
    }
    return result;
}

Java

public int[] twoSum(int[] numbers, int target) {
    int[] result = new int[2];
    Map map = new HashMap();
    for (int i = 0; i < numbers.length; i++) {
        if (map.containsKey(target - numbers[i])) {
            result[1] = i + 1;
            result[0] = map.get(target - numbers[i]);
            return result;
        }
        map.put(numbers[i], i + 1);
    }
    return result;
}

Maybe The Shorest Solution

public int[] twoSum(int[] nums, int target) {
    HashMap map = new HashMap<>();
    for(int i = 0; i < nums.length; i++){
        if(map.containsKey(target - nums[i])) 
            return new int[] {map.get(target - nums[i]) + 1, i + 1};
        else map.put(nums[i], i);
        }
    return null;
    }

为什么要用Hashmap？

Hashmap根据键的hashCode值存储数据，大多数情况下可以直接定位到它的值，因而具有很快的访问速度，为了将时间复杂度降到O(n)，我们使用了Hashmap。

methods of Hashmap

• HashMap()
  构造一个具有默认初始容量 (16) 和默认加载因子 (0.75) 的空 HashMap。
• HashMap(int initialCapacity)
  构造一个带指定初始容量和默认加载因子 (0.75) 的空 HashMap
• HashMap(int initialCapacity, float loadFactor)
  构造一个带指定初始容量和加载因子的空 HashMap
• void clear()
  从此映射中移除所有映射关系。
• boolean containsKey(Object key)
  如果此映射包含对于指定的键的映射关系，则返回 true。
• boolean containsValue(Object value)
  如果此映射将一个或多个键映射到指定值，则返回 true。
• V put(K key, V value)
  在此映射中关联指定值与指定键。-

在HashMap中通过get()来获取value，通过put()来插入value，ContainsKey()则用来检验对象是否已经存在。可以看出，和ArrayList的操作相比，HashMap除了通过key索引其内容之外，别的方面差异并不大。

使用HashMap后，运行速度相比于原来的275ms确实有明显提升

Runtime: 8ms
Your runtime beats 56.48% of java submissions.