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LeetCode Crack Note --- 1. Two Sum

2016-03-01 11:37 344 查看

Discription

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].

Difficulty: ★

Solving Strategy

将原数组深拷贝一份，并进行排序，得到数组nums2；

使用两个指针，分别指向nums2的首尾，两个指针同时向中间移动，并判断当前下标对应元素的和是否等于target。（这点在特定情况下，可以节省大量时间，eg. nums=[0,1,2,3,…,9999]，target=1000）

My Solution

1 # Use Python3
2 class Solution:
3     def twoSum(self, nums, target):
4         """
5         :type nums: List[int]
6         :type target: int:rtype: List[int]
7         """
8
9         nums2 = nums[:] # Copy a new list for sorting
10         nums2.sort()
11         ii = 0
12         count = nums.__len__()
13         jj = count-1
14
15         val1 = 0
16         val2 = 0
17         while ii < jj:
18             if nums2[ii] + nums2[jj] == target:
19                 val1 = nums2[ii]
20                 val2 = nums2[jj]
21                 break
22             elif nums2[ii] + nums2[jj] < target:
23                 ii = ii + 1
24             elif nums2[ii] + nums2[jj] > target:
25                 jj = jj - 1
26
27         indexList = []
28         for i in range(count):
29             if nums[i] == val1:
30                 indexList.append(i)
31                 break
32
33         for j in range(count-1, -1, -1):
34             if nums[j] == val2:
35                 indexList.append(j)
36                 break
37
38         indexList.sort()
39         return [indexList[0], indexList[1]]

Runtime: 85 ms

Official Solution

Approach #1 (Brute Force) [Accepted]

The brute force approach is simple. Loop through each element x and find if there is another value that equals to target - x.

1 public int[] twoSum(int[] nums, int target) {
2     for (int i = 0; i < nums.length; i++) {
3         for (int j = i + 1; j < nums.length; j++) {
4             if (nums[j] == target - nums[i]) {
5                 return new int[] { i, j };
6             }
7         }
8     }
9     throw new IllegalArgumentException("No two sum solution");
10 }

Complexity Analysis:

Time complexity : O(n). We traverse the list containing n elements exactly twice. Since the hash table reduces the look up time to O(1), the time complexity is O(n).

Space complexity : O(n). The extra space required depends on the number of items stored in the hash table, which stores exactly n elements.

Approach #2 (Two-pass Hash Table) [Accepted]

Approach #3 (One-pass Hash Table) [Accepted]

Reference

[南郭子綦's blog]

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