LeetCode刷题Day1 Two Sum

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 thesame element twice.

Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

我的做法：

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        list1= []
        for i in range(0,len(nums)):
            for j in range(i+1,len(nums)):
                if (nums[i]+nums[j]== target):
                    list1.append(i)
                    list1.append(j)
                    return list1


测试样例：

INPUT

[3,2,4]6

YOUR ANSWER && EXPECTED ANSWER

[1,2]

      

再看别人已经写的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(n2)O(n^2) For each element, we try to find its complement by looping through the rest of array which takesO(n)O(n)time. Therefore, the time complexity isO(n2)O(n^2)

• Space complexity : O(1)O(1)

这种暴力破解方法就是我写的方法

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

To improve our run time complexity, we need a more efficient way to check if the complement exists in the array. If the complement exists, we need to look up its index. What is the best way to maintain a mapping of each element in the array to its index? A hash table.

We reduce the look up time from $O(n)$ to $O(1)$ by trading space for speed. A hash table is built exactly for this purpose, it supports fast look up innear constant time. I say "near" because if a collision occurred, a look up could degenerate to$O(n)$ time. But look up in hash table should be amortized $O(1)$ time as long as the hash function was chosen carefully.

A simple implementation uses two iterations. In the first iteration, we add each element's value and its index to the table. Then, in the second iteration we check if each element's complement ($target-nums[i]$) exists in the table. Beware that the complement must not be $nums[i]$ itself!

public int[] twoSum(int[] nums, int target) {    Map<Integer, Integer> map = new HashMap<>();    for (int i = 0; i < nums.length; i++) {        map.put(nums[i], i);    }    for (int i = 0; i < nums.length; i++) {        int complement = target - nums[i];        if (map.containsKey(complement) && map.get(complement) != i) { //map.get 得到complement对应的index            return new int[] { i, map.get(complement) };        }    }    throw new IllegalArgumentException("No two sum solution");}

Complexity Analysis:

• Time complexity : O(n)O(n). We traverse the list containingnn elements exactly twice. Since the hash table reduces the look up time toO(1)O(1) the time complexity isO(n)O(n).

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

收获：终于知道怎么用Hash Table了

• 
  

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Approach #3 (One-pass Hash Table) [Accepted] 确实是奇思妙想

It turns out we can do it in one-pass. While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.

public int[] twoSum(int[] nums, int target) {    Map<Integer, Integer> map = new HashMap<>();    for (int i = 0; i < nums.length; i++) {        int complement = target - nums[i];        if (map.containsKey(complement)) {            return new int[] { map.get(complement), i };        }        map.put(nums[i], i);    }    throw new IllegalArgumentException("No two sum solution");}

Complexity Analysis:

• Time complexity : O(n)O(n). We traverse the list containing$n$ elements only once. Each look up in the table costs only O(1)O(1) time.

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