[LintCode] Two Sum

最新推荐文章于 2025-12-20 16:06:47 发布

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原文链接：http://www.cnblogs.com/lz87/p/7203577.html

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#runtime

本文解析了在整数数组中寻找两个数使它们的和等于特定目标值的问题。提供了三种解决方案：使用哈希表的两次遍历(O(n)时间和空间复杂度)，一次遍历的改进版以及利用排序和双指针技术实现O(n log n)时间复杂度的方法。

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are NOT zero-based.

You may assume that each input would have exactly one solution

Example

numbers=[2, 7, 11, 15], target=9

return [1, 2]

Challenge

Either of the following solutions are acceptable:

O(n) Space, O(nlogn) Time
O(n) Space, O(n) Time

Solution 1. O(n) runtime, O(n) space using hash map, two pass

 1 public class Solution {
 2     public int[] twoSum(int[] numbers, int target) {
 3         int[] result = new int[2];
 4         if(numbers == null || numbers.length <= 1)
 5         {
 6             return result;
 7         }
 8         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
 9         for(int i = 0; i < numbers.length; i++)
10         {
11             map.put(numbers[i], i);
12         }
13         for(int i = 0; i < numbers.length; i++)
14         {
15             if(map.containsKey(target - numbers[i]))
16             {
17                 result[0] = i + 1;
18                 result[1] = map.get(target - numbers[i]) + 1;
19                 break;
20             }
21         }
22         return result;
23     }
24 }

Solution 2. O(n) runtime, O(n) space using hash map, one pass

 1 public class Solution {
 2     public int[] twoSum(int[] numbers, int target) {
 3         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();       
 4         for(int i = 0; i < numbers.length; i++)
 5         {
 6             if(map.get(numbers[i]) != null)
 7             {
 8                 int[] result = {map.get(numbers[i]) + 1, i + 1};
 9                 return result;
10             }
11             map.put(target - numbers[i], i);
12         }
13         int[] result = {};
14         return result;
15     }
16 }

Solution 3. O(n*logn) runtime, O(n) space using sort and two pointers

 1 class ArrayEntry
 2 {
 3     protected int val;
 4     protected int index;
 5     public ArrayEntry(int val, int index)
 6     {
 7         this.val = val;
 8         this.index = index;
 9     }
10 }
11 public class Solution {
12     public int[] twoSum(int[] numbers, int target) {
13         int[] result = new int[2];
14         if(numbers == null || numbers.length <= 1)
15         {
16             return result;
17         }
18         ArrayEntry[] entries = new ArrayEntry[numbers.length];
19         for(int i = 0; i < numbers.length; i++)
20         {
21             entries[i] = new ArrayEntry(numbers[i], i);
22         }
23         Arrays.sort(entries, new Comparator<ArrayEntry>()
24             {
25                 public int compare(ArrayEntry a1, ArrayEntry a2)
26                 {
27                     return Integer.compare(a1.val, a2.val);
28                 }
29             }
30         );
31         int i = 0, j = entries.length - 1;
32         while(i < j)
33         {
34             int sum = entries[i].val + entries[j].val;
35             if(sum == target)
36             {
37                 if(entries[i].index < entries[j].index)
38                 {
39                     result[0] = entries[i].index + 1;
40                     result[1] = entries[j].index + 1;
41                 }
42                 else
43                 {
44                     result[0] = entries[j].index + 1;
45                     result[1] = entries[i].index + 1;
46                 }
47                 break;
48             }
49             else if(sum < target)
50             {
51                 i++;
52             }
53             else
54             {
55                 j--;
56             }
57         }
58         return result;
59     }
60 }