来自专栏 · LeetCode Description Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: The number of elements initialized in nums1 and nums2 are m and n respectively.You may assume that nums1 has enough space (size that is greater or equal...
publicstaticvoidmain(String[] args){ Easy_088_MergeSortedArray instance =newEasy_088_MergeSortedArray();int[] nums1 = {1,2,2,3,4,5,0,0,0};intm =6;int[] nums2 = {2,5,8};intn =3;longstart = System.nanoTime(); instance.merge(nums1, m, nums2, n);longend = System.nano...
题目: 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 tom+n) to hold additional elements from B. The number of elements initialized in A and B aremandnrespectively. 题解: 这...
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 按照归并排序的惯性思...
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. """ # i/j 分别表示 nums1/nums2 中还未使用的最大数的下标 i, j = m - 1, n - 1 # k 表示 nums1 中下一个该...
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 initialized in nums1and nums2 ...
Forjava, it's difficult to extend array, thus difficult to construct a new slot to save infinity2147483647. We can usemodulusto accomplish this target, as long as we set the visited slot to be infinity. Running time isO(m+n). classSolution{publicvoidmerge(int[]nums1,intm,int[]nums2,...
So running it recursively did improve the runtime A LOT. I'm gonna just let go of the problem left with mergeTwoLists, since it wouldn't affect the runtime but just created more space in the memory. Oh almost forgot, here's my final solution, I'm actually pretty proud of solving th...
public class Solution { public void merge(int[] nums1, int m, int[] nums2, int n) { m = m - 1; n = n - 1; int i = m + n + 1; while(m >= 0 || n >= 0){ if(m < 0){ nums1[i--] = nums2[n--];
java: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{publicvoidmerge(int[]nums1,int m,int[]nums2,int n){int i=m-1,j=n-1,k=m+n-1;while(i>=0&&j>=0)nums1[k--]=nums1[i]>=nums2[j]?nums1[i--]:nums2[j--];while(j>=0)nums1[k--]=nums2[j--];}}...