1.two sum

Question:

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

 

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

 

Constraints:

• 2 <= nums.length <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

Answer:

1. Run time:4543 ms 

   Memory:14.1 M

　　class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range (len(nums)-1):
            j = i + 1
            while j < len(nums):
                if nums[i] + nums[j] == target:
                    return [i,j]
                j += 1
                
a=Solution()
print(a.twoSum([2,7,11,15],9))
print(a.twoSum([3,2,4],6))
print(a.twoSum([3,3],6))

2. Run time:56 ms 

   Memory:14.1 M'

class Solution():
    def twoSum(self, nums, target):
        values = {}
        for i, num in enumerate(nums):
            remaining = target - num
            if remaining in values:
                return [i, values[remaining]]
            else:
                values[num] = i

　　

class Solution():    def twoSum(self, nums, target):        values = {}        for i, num in enumerate(nums):            remaining = target - num            if remaining in values:                return [i, values[remaining]]            else:                values[num] = i