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Leetcode —— two_sum

问题描述

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:

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

Example 2:

<strong>Input:</strong> nums = [3,2,4], target = 6
<strong>Output:</strong> [1,2]

Example 3:

<strong>Input:</strong> nums = [3,3], target = 6
<strong>Output:</strong> [0,1]

Constraints:

  • 2 <= nums.length <= 10<sup>3</sup>
  • -10<sup>9</sup> <= nums[i] <= 10<sup>9</sup>
  • -10<sup>9</sup> <= target <= 10<sup>9</sup>
  • Only one valid answer exists.

自己写的暴力解法

def two_sum(nums, target)
  for i in 0...nums.size
    for j in (i+1)...nums.size
      if nums[i] + nums[j] == target
        puts i, j
        return [i, j]
      end
    end
  end
end

虽然能用,但是不优雅

别人家的优雅解法

def two_sum(nums, target)
  map = {}

  nums.each_with_index { |val, index|
    n = target - val
    if map.include?(n)
      return [map[n], index]
    else
        map[val] = index
    end
  }
end

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