LeetCode Js-1929. Concatenation of Array

    AllenLi
    LeetCode Js-Array

    Given an integer array nums of length n, you want to create an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed). Specifically, ans is the concatenation of two nums arrays. Return the array ans. 

    給予一個整數陣列 nums 和長度為 n，你想新增一個長度為 2n 的陣列 ans。
其中 ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed).
具體來說，ans 是兩次 nums 陣列的串聯。
回傳 ans 的陣列。

    Example 1:

    Input: nums = [1,2,1]
Output: [1,2,1,1,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]]
- ans = [1,2,1,1,2,1]

    Example 2:

    Input: nums = [1,3,2,1]
Output: [1,3,2,1,1,3,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]]
- ans = [1,3,2,1,1,3,2,1]

    Solution:
    1. 建立 ans = []
    2. 將 for(i < 2) 迴圈執行兩次 3. 將 for(j < nums.length)將每個 nums 的元素執行一次，並運用 .push() 放入 ans。 4. 回傳 ans。 Code.1: For Loop (BigO(2n)

    var getConcatenation = function(nums) {
    let array = []
    for (let i = 0; i < 2; i++) {
        for (let j = 0; j < nums.length; j++) {
            array.push(nums[j])
        }
    }
    return array;
}

    FlowChart:
    Example 1

    Input: nums = [1,2,1]

step.1 
i = 0
ans.push(nums[0]) => [1]
ans.push(nums[1]) => [1, 2]
ans.push(nums[2]) => [1, 2, 1]

step.2
i = 1
ans.push(nums[0]) => [1, 2, 1, 1]
ans.push(nums[1]) => [1, 2, 1, 1, 2]
ans.push(nums[2]) => [1, 2, 1, 1, 2, 1]

return ans //[1, 2, 1, 1, 2, 1]

    Example 2

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

step.1 
i = 0
ans.push(nums[0]) => [1]
ans.push(nums[1]) => [1, 3]
ans.push(nums[2]) => [1, 3, 2]
ans.push(nums[3]) => [1, 3, 2, 1]

step.2
i = 1
ans.push(nums[0]) => [1, 3, 2, 1, 1]
ans.push(nums[1]) => [1, 3, 2, 1, 1, 3]
ans.push(nums[2]) => [1, 3, 2, 1, 1, 3, 2]
ans.push(nums[3]) => [1, 3, 2, 1, 1, 3, 2, 1]

return and //[1, 3, 2, 1, 1, 3, 2]

    Code.2: For Loop (BigO(n))

    var getConcatenation = function(nums) {
  const length = nums.length
  for (let i = 0; i < length; i++) {
    nums.push(nums[i])
  }
  return nums;
}

    Code.3: Concat (BigO(n))

    var getConcatenation = function(nums) {
  return nums.concat(nums) 
}

    .concat()

    .concat()：合併兩個或多個陣列。

const array1 = ['a', 'b', 'c'];
const array2 = ['d', 'e', 'f'];
const array3 = array1.concat(array2);

console.log(array3);
// expected output: Array ["a", "b", "c", "d", "e", "f"]

    Code.4: split (BigO(n))

    var getConcatenation = function(nums) {
  s = nums.join(",")
  ss = s + "," + s
  return ss.split(",");
}

    Code.5: ...array (BigO(n))

    var getConcatenation = function(nums) {
  return [...nums, ...nums];
}

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