LeetCode Js-88. Merge Sorted Array

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1.
To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored.
nums2 has a length of n.

你給予兩個整數陣列 nums1 和 nums2，兩個整數 m 和 n 分別代表 nums1 和 nums2 的元素個數。
合併 nums1 和 nums2 成為一個「非遞減」的訂單排序，不過是新增到 nums1 中。
關於這個狀況，nums1 有一個長度是 m + n，而前面的 m 元素表示為被合併，後面的 n 元素是變為 0 並忽略。
ex.
nums1 = [1, 2, 3, 0, 0, 0], nums2 = [1, 2, 3]
Output = [1, 1, 2, 2, 3, 3]

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.

Solution:
1. 宣告 index = 0
2. 使用 for 迴圈將 nums2 的值合併到nums1。
3. 雙 for 迴圈進行兩兩比對，當 nums1 > nums2 時，兩兩數字互換。

Code 1: BigO(n^2)

var merge = function(nums1, m, nums2, n) {
  // let the index become the starting point
  let index = 0;

  // merge in reverse order
  for (let i = m; i < (m + n); i++) {
    nums1[i] = nums2[index];
    index++;
  }

  // use destructuring assignment to swap variables.
  for (let j = 0; j < nums1.length - 1; j++) {
    for (let k = j + 1; k < nums1.length; k++) {
      if (nums1[j] > nums1[k]) {
        [nums1[j], nums1[k]] = [nums1[k], nums1[j]];
      }
    }
  }
};

FlowChart:
Example 1

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3

step.1 let i = m //3
i < (3 + 3) //3 < 6
nums1[3] = nums2[0] //[1,2,3,0,0,0] => [1,2,3,0,0,0]
                                              2
index++ //1

i++ //4
i < (3 + 3) //4 < 6
nums1[4] = nums2[1] //[1,2,3,2,0,0] => [1,2,3,2,0,0]
                                                5
index++ //2

i++ //5
i < (3 + 3) //5 < 6
nums1[5] = nums2[2] //[1,2,3,2,5,0] => [1,2,3,2,5,0]
                                                  6
index++ //3

step.2 nums1 = [1,2,3,2,5,6] 
j = 0, j < 5
k = 0 + 1, k < 6 ...  //k = 1, 2, 3, 4 ,5
nums1[0] > nums1[1] //1 < 2
nums1[0] > nums1[2] //1 < 3
nums1[0] > nums1[3] //1 < 2
nums1[0] > nums1[4] //1 < 5
nums1[0] > nums1[5] //1 < 6

j = 1
k = 1 + 1, k < 6 ...  //k = 1, 2, 3, 4 ,5
nums1[1] > nums1[2] //2 < 3
nums1[1] > nums1[3] //2 < 2
nums1[1] > nums1[4] //2 < 5
nums1[1] > nums1[5] //2 < 6

j = 2
k = 2 + 1, k < 6 ...  //k = 1, 2, 3, 4 ,5
nums1[2] > nums1[3] //3 < 2
box = nums1[2] = 3; nums1[2] = nums1[3] => 2; nums1[3] = box => 3
nums1 = [1,2,2,3,5,6] 
nums1[2] > nums1[4] //2 < 5
nums1[2] > nums1[5] //2 < 6

j = 3
k = 3 + 1, k < 6 ...  //k = 1, 2, 3, 4 ,5
nums1[3] > nums1[4] //3 < 5
nums1[3] > nums1[5] //3 < 6

j = 4
k = 4 + 1, k < 6 ...  //k = 1, 2, 3, 4 ,5
nums1[4] > nums1[5] //5 < 6

nums1 = [1,2,2,3,5,6] 

Example 2

Input: nums1 = [1], m = 1, nums2 = [], n = 0

step.1 let i = m //1
i < (1 + 0) //1 <>1
nums1[1] = nums2[0] //[1] => [1]

step.2 nums1 = [1] 
j = 0, j < 0
k = 0 + 1, k < 1 ...  //k = 1

Example 3

Input: nums1 = [0], m = 0, nums2 = [1], n = 1

step.1 let i = m //0
i < (0 + 1) //1
nums1[0] = nums2[0] //[0] => [1]

step.2 nums1 = [1] 
j = 0, j < 0
k = 0 + 1, k < 1 ...  //k = 1

nums1[0] = [1]

Code 2: BigO(n log n)

var merge = function (nums1, m, nums2, n) {
  for (let i = 0; i < n; i++) {
    nums1[m + i] = nums2[i];
  }

  return nums1.sort((a, b) => a - b);
};

Code 3: BigO(n log n)

var merge = function (nums1, m, nums2, n) {
  nums1.length = m
  nums2.forEach(n => nums1.push(n));
  return nums1.sort((a, b) => a-b);
};

Code 4: BigO(n log n)

var merge = function(nums1, m, nums2, n) {
  nums1.splice(m, n, ...nums2);
  return nums1.sort((a, b) => a - b);
};

Code 5: BigO(n)

var merge = function(nums1, m, nums2, n) {
    // let index of completed nums1
    let k = m + n - 1

	// merge in reverse order
    while (m > 0 && n > 0) {
        if (nums1[m - 1] > nums2[n - 1]) {
            nums1[k] = nums1[m - 1]
            m--
        } else {
            nums1[k] = nums2[n - 1]
            n--
        }
        k--
    }

    // if m = 0 then push nums2 in nums1
    while (n > 0) {
        nums1[k] = nums2[n - 1]
        k--
        n--
    }
};