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LeetCode 01 - Two Sum

  • Difficulty: Easy.
  • Related Topics: Array, Hash Table.
  • Link: leetcode

Problem​

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].

Constraints:​

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

Solution 1: two head​

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */

var twoSum = function (nums, target) {
  // only one result
  if (nums.length === 2) {
    return [0, 1];
  }

  for (let i = 0; i < nums.length - 1; i++) {
    for (let j = 1; j < nums.length; j++) {
      if (nums[i] + nums[j] === target && i !== j) {
        return [i, j];
      }
    }
  }
};

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

Solution 2: hash map​

var twoSum = function (nums, target) {
  let obj = {};
  for (let i = 0; i < nums.length; i++) {
    if (target - nums[i] in obj) {
      return [obj[target - nums[i]], i];
    } else {
      obj[nums[i]] = i;
    }
  }
};

Solution 3: one head solution​

var twoSum = function (nums, target) {
  for (let i = 0; i < nums.length; i++) {
    let index = nums.indexOf(target - nums[i]);
    if (index !== -1 && i !== index) {
      return [i, index];
    }
  }
};