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]