(关注第二次while循环。) defmerge_sorted(nums1,m,nums2,n):pointer_1=m-1pointer_2=n-1pointer=m+n-1whilepointer_2>=0andpointer_1>=0:ifnums1[pointer_1]>nums2[pointer_2]:nums1[pointer]=nums1[pointer_1]pointer_1-=1else:nums1[pointer]=nums2[pointer_2]pointer_2-=1pointer-=1while...
88. Merge Sorted Array 思路一:把nums2直接复制到nums1中,然后sort,但是你会发现地址在sort时返回的地址已经发生改变,所以这种解法是不对的。 classSolution:defmerge1(self,nums1,m,nums2,n):print(id(nums1))len1=len(nums1)len2=nforiinrange(len1-len2,len1):nums1[i]=nums2[i-(len1-len2)...
The final sorted array should not be returned by the function, but instead bestored inside the arraynums1. To accommodate this,nums1has a length ofm + n, where the firstmelements denote the elements that should be merged, and the lastnelements are set to0and should be ignored.nums2has ...
leetcode 88 Merge Sorted Array 归并排序 归并排序:先将数组一分为二,将左边部分排序(同样将其一分为二),再将右边部分排序,最后逐层归并。(分治策略)(稳定排序)。 算法稳定性 -- 假设在数列中存在a[i]=a[j],若在排序之前,a[i]在a[j]前面;并且排序之后,a[i]仍然在a[j]前面。则这个排序算法是稳定...
Merge Sorted Array Given two sorted integer arrays A and B, merge B into A as one sorted array. Note: You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m andn ...
【Leetcode】Merge Sorted Array https://leetcode.com/problems/merge-sorted-array/ 题目: nums1 and nums2, merge nums2 into nums1 Note: You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements ...
Given two sorted integer arrays A and B, merge B into A as one sorted array. Note: You may assume that A has enough space to hold additional elements from B. The number of elements initialized in A and B are m and n respectively. 题目大意 给定两个排序的整型数组A和B,将B合并到A中。
Merge Sorted Array. [easy] 链接: Merge Sorted Array 题意:混合插入有序数组。 分析: 一开始没看懂题意,就先把两个数组合并再输出,没有注意合并后的数组的长度为m+n; 可以从后往前附值,比较nums1和nums2最后一个元素的大小,把大的那一个放入m+n-1的位置中, 以此类推。如果nums1中所有元素都比nums2...
So, while merging, we don’t need an auxiliary array to store the elements in the correct order; instead, we can remove the head node from a list, append it to the end of the sorted list and move the pointer to the next node using “next” pointer (if possible). Therefore, by exp...