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At least one question a day ： An integral +10 branch
If you do one more question （ Or one more way to answer ）, Then the points of the day +20 branch （+10+10）
If you do more than three , Start with question 3 +20 branch （ Such as ： If you do three questions, you will score -10+10+20=40; If you do four questions, you will score –10+10+20+20=60）Initial classification 100 branch
If you haven't done the problem for one day , Then deduct points -10 branch （ Saturday 、 Except Sunday ： rest ）
insist ！！！
Here are two buttons Non decreasing order Array of arranged integers nums1 and nums2, There are two other integers m and n , respectively nums1 and nums2 The number of elements in .
Would you please Merge nums2 To nums1 in , Make the merged array press Non decreasing order array .
Be careful ： Final , The merged array should not be returned by the function , It's stored in an array nums1 in . In response to this situation ,nums1 The initial length of is m + n, The top m Elements represent the elements that should be merged , after n Elements are 0 , It should be ignored .nums2 The length of is 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]
explain ： Need merger [1,2,3] and [2,5,6] .
The combined result is [1,2,2,3,5,6] , In which, bold italics indicates nums1 The elements in .
Example 2：
Input ：nums1 = [1], m = 1, nums2 = [], n = 0
Output ：[1]
explain ： Need merger [1] and [] .
The combined result is [1] .
Example 3：
Input ：nums1 = [0], m = 0, nums2 = [1], n = 1
Output ：[1]
explain ： The array to be merged is [] and [1] .
The combined result is [1] .
Be careful , because m = 0 , therefore nums1 No elements in .nums1 The only remaining 0 Just to ensure that the merged results can be successfully stored in nums1 in .
analysis ：
The subject is very simple , We know the length that needs to be merged , data , Just replace the... In the first array 0, Then sort the original array .
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. """ if m ==0: nums1[:] = nums2[:] else: # section nums1[:] = nums1[:m] for i in range(n): nums1.append(nums2[i]) nums1.sort()
Although a little opportunistic , however , Is that so? , EH ！~
If we're a little more clever , It can also be sliced directly ！！！
Go straight to the code ！！！ Ha ha ha
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. """ nums1[m:] = nums2[:] nums1.sort()
Two lines of code to do it directly , Let's stop here today , Reading papers , Projects o(╥﹏╥)o
author ： Power button (LeetCode)
link ：https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/xnumcr/
source ： Power button （LeetCode）
Score today ：+10+10
Total score ：730come on. ！！！
