5/6每日一題

 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.

 

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.

class Solution:
  def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
      """
      Do not return anything, modify nums1 in-place instead.
      """
      p1=m-1 #從最後開始比較
      p2=n-1
      in_dex=m+n-1 #合併所有元素的索引範圍
      if p1<0:  #當nums1不存在元素時
          nums1[:]=nums2[:]
      
      
      while p1 >=0 and p2 >=0:  #在檢查完所有元素之前持續動作
        if nums1[p1] > nums2 [p2]:
          nums1[in_dex]=nums1[p1] #最大在最後          
          p1 -=1                  #指針向左移動    
        else:
          nums1[in_dex]=nums2[p2]
          p2 -= 1                 #指針向左移動
        in_dex -= 1
      
      
      if p2 >=0:  #檢查是否排序完畢  
        nums1[:p2+1]=nums2[:p2+1]  
        p2-=1
                     
      return nums1