961. N-Repeated Element in Size 2N Array

Easy

題目描述

Description

You are given an integer array nums with the following properties:

• nums.length == 2 * n.
• nums contains n + 1 unique elements.
• Exactly one element of nums is repeated n times.

Return the element that is repeated n times.

Example 1:

Input: nums = [1,2,3,3]
Output: 3

Example 2:

Input: nums = [2,1,2,5,3,2]
Output: 2

Example 3:

Input: nums = [5,1,5,2,5,3,5,4]
Output: 5

Constraints:

• 2 <= n <= 5000
• nums.length == 2 * n
• 0 <= nums[i] <= 104
• nums contains n + 1 unique elements and one of them is repeated exactly n times.

解答

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var repeatedNTimes = function (nums) {
  const counter = {};

  for (let i = 0; i, nums.length; i++) {
    if (counter[nums[i]] === 1) {
      return nums[i];
    } else {
      counter[nums[i]] = 1;
    }
  }
};

解題思路

根據題目描述只會有一個 element 重複，所以先用一個 counter 紀錄出現的數字，當出現重複就直接 return

心得

直接把直覺轉成 JS 程式碼竟然一次過耶! 謝謝哩扣 🥳