Two Sum

Example:

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

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

 1 #include <bits/stdc++.h>
 2 #include <map>
 3 #include <vector>
 4 using namespace std;
 5 
 6 class Solution
 7 {
 8   public:
 9     vector<int> twoSum(vector<int> &numbers, int target)
10     {
11         unordered_map<int, int> hash;
12         vector<int> result;
13         for (int i = 0; i < numbers.size(); i++)
14         {
15             int numberToFind = target - numbers[i];
16             if (hash.find(numberToFind) != hash.end())
17             {
18                 result.push_back(hash[numberToFind] + 1);
19                 result.push_back(i + 1);
20                 return result;
21             }
22             hash[numbers[i]] = i;
23         }
24         return result;
25     }
26 };
27 int main()
28 {
29     vector<int> v;
30     int target;
31     int num;
32     while (cin >> num)
33     {
34         if (num == 0)
35             break;
36         v.push_back(num);
37     }
38     cin >> target;
39     Solution sol;
40     vector<int> result;
41     result = sol.twoSum(v, target);
42     cout << result[0] << " " << result[1] << endl;
43     return 0;
44 }

 1 import java.util.ArrayList;
 2 import java.util.HashMap;
 3 import java.util.Map;
 4 import java.util.Scanner;
 5 
 6 class Solution {
 7 
 8     public int[] twoSum(int[] numbers, int target) {
 9         int[] result = new int[2];
10         Map<Integer, Integer> map = new HashMap<>(); //其中用到互加性
11         for (int i = 0; i < numbers.length; i++) {
12             if (map.containsKey(target - numbers[i])) {    //通过map检查是否存在符合的值，若存在将索引赋给result[1]
13                 result[1] = i;
14                 result[0] = map.get(target - numbers[i]);
15                 return result;
16             }
17             map.put(numbers[i], i);
18         }
19         return result;
20     }
21 }
22 
23 public class TwoSum {
24     public static void main(String[] args) {
25         Scanner input = new Scanner(System.in);
26         ArrayList<Integer> array = new ArrayList<>();
27         while (!input.hasNext("0")) {
28             array.add(Integer.valueOf(input.next()));
29         }
30         Scanner read = new Scanner(System.in);
31         Integer target = read.nextInt();
32         int[] d = new int[array.size()];
33         for (int i = 0; i < array.size(); i++) {
34             d[i] = array.get(i);
35         }
36         Solution sol = new Solution();
37         int[] result = sol.twoSum(d, target);
38         System.out.printf("%d	 %d", result[0], result[1]);
39 
40     }
41 }

调用了getNode(hash(key), key)方法，tab[(n - 1) & hash]执行了如下操作：
1. 指针first指向那一行数组的引用（那一行数组是通过table下标范围n-1和key的hash值计算出来的），若命中，则通过下标访问数组，时间复杂度为O(1)
2. 如果没有直接命中（key进行hash时，产生相同的位运算值），存储方式变为红黑树，那么遍历树的时间复杂度为O(n)。

/**
     * Implements Map.get and related methods
     *
     * @param hash hash for key
     * @param key the key
     * @return the node, or null if none
     */
    final Node<K,V> getNode(int hash, Object key) {
        Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (first = tab[(n - 1) & hash]) != null) {
            // 直接命中
            if (first.hash == hash && // always check first node
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            // 未命中
            if ((e = first.next) != null) {
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }