Viiisit [LeetCode] - 80. Remove Duplicates from Sorted Array II

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Problem - Remove Duplicates from Sorted Array II

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

翻譯蒟蒻

移除 nums 陣列中的相同元素，並將其中重複的元素最多保留兩次，並且保持相對順序，返回最終的數量。(要求在不使用額外內存空間（即不建立一個新的陣列）的情況下，對現有的進行修改)

• Example 1:

  Input: nums = [1,1,1,2,2,3]
  Output: 5, nums = [1,1,2,2,3,_]
  Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
  It does not matter what you leave beyond the returned k (hence they are underscores).

• Example 2:

  Input: nums = [0,0,1,1,1,1,2,3,3]
  Output: 7, nums = [0,0,1,1,2,3,3,_,_]
  Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
  It does not matter what you leave beyond the returned k (hence they are underscores).

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Solution - JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
  // 一開始的兩個元素一定會被記錄
  let position = 2; 
  
  // 從陣列第三個元素開始檢查，如果不同就保留，確保每個元素最多就兩個
  for (let i = 2; i < nums.length; i++) {
    if (nums[i] !== nums[position - 2]) {
      nums[position] = nums[i];
      position++;
    }
  }
  
  // 表示處理過後的 Array 長度
  return position;
};

Solution - Ruby

def remove_duplicates(nums)
  return nums.length if nums.length <= 2

  position = 2

  for i in 2..(nums.length - 1)
    if nums[i] != nums[position - 2]
      nums[position] = nums[i]
      position += 1
    end
  end

  position
end

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