LeetCode Crack Note --- 1. Two Sum

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

1. 将原数组深拷贝一份，并进行排序，得到数组nums2；
2. 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

 # Use Python3
 class Solution:
     def twoSum(self, nums, target):
         """
         :type nums: List[int]
         :type target: int:rtype: List[int]
         """

         nums2 = nums[:] # Copy a new list for sorting
         nums2.sort()
         ii = 0
         count = nums.__len__()
         jj = count-1

         val1 = 0
         val2 = 0
         while ii < jj:
             if nums2[ii] + nums2[jj] == target:
                 val1 = nums2[ii]
                 val2 = nums2[jj]
                 break
             elif nums2[ii] + nums2[jj] < target:
                 ii = ii + 1
             elif nums2[ii] + nums2[jj] > target:
                 jj = jj - 1

         indexList = []
         for i in range(count):
             if nums[i] == val1:
                 indexList.append(i)
                 break

         for j in range(count-1, -1, -1):
             if nums[j] == val2:
                 indexList.append(j)
                 break 

         indexList.sort()
         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.

 public int[] twoSum(int[] nums, int target) {
     for (int i = 0; i < nums.length; i++) {
         for (int j = i + 1; j < nums.length; j++) {
             if (nums[j] == target - nums[i]) {
                 return new int[] { i, j };
             }
         }
     }
     throw new IllegalArgumentException("No two sum solution");
 }

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]