11.6 Sorts - Merge Sort Code 归并排序代码是【生肉】圣地亚哥州立大学 -数据结构与算法 - Data Structures and Algorithms的第82集视频,该合集共计89集,视频收藏或关注UP主,及时了解更多相关视频内容。
这道题目其实和基本排序算法中的 merge sort 非常像。我们先来回顾一下 merge sort 的 merge 过程。merge 的过程 可以 是先比较两个数组的头元素，然后将较小的推到最终的数组中，并将其从原数组中出队列。不断循环直到两个数组都为空。具体代码如下：// 将 nums1 和 nums2 合并function merge(nums1, nums...
Merge step Writing the Code for Merge Algorithm A noticeable difference between the merging step we described above and the one we use for merge sort is that we only perform the merge function on consecutive sub-arrays. This is why we only need the array, the first position, the last index...
for(intj =0; j < help.length; j++) { arr[left+j] = help[j]; } returnresult; } publicstaticvoidmain(String[] args) { Code02_SmallSum sort =newCode02_SmallSum(); int[] arr = {5,2,3,4,1}; //int[] arr = {1,2,3,4}; intsmallSum =sort.process(arr,0, arr.length-1)...
void mergeSortedArray(int A[], int m, int B[], int n) { // write your code here int *res=new int[m+n]; for(int i=0;i<m;i++){ res[i]=A[i]; } for(int i=m;i<m+n;i++){ res[i]=B[i-m]; } sort(res,res+m+n);// ...
publicvoidmerge(int[] nums1,intm,int[] nums2,intn){for(inti=m, k=0; i<nums1.length && k < n ; i++,k++) { nums1[i] = nums2[k]; }for(intj=0; j<nums1.length-1; j++) {for(inth=0; h<nums1.length-1-j; h++) {if(nums1[h] > nums1[h+1]) {inttemp=nums1...
class Solution: def merge_sort(self, nums): # If the array length is less than or equal to 1, return the array (base case for recursion) if len(nums) <= 1: return nums # Find the middle point mid = len(nums) // 2 # Recursively sort the left half left_half = self.merge_sort...
int[][] intervals){List<int[]> result =newArrayList<>();// Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));Arrays.sort(intervals,newComparator<int[]>(){publicintcompare(int[] a1,int[] a2){return a1[]- a2[];}});inthead= intervals[][], tail = intervals[][1];for(...
classSolution{publicvoidmerge(int[]nums1,int m,int[]nums2,int n){for(int i=0;i!=n;++i){nums1[m+i]=nums2[i];}Arrays.sort(nums1);}} 3、时间复杂度 时间复杂度 :O((m+n)log(m+n)) 排序序列长度为m+n,套用快排的时间复杂度即可,也就是O((m+n)log(m+n)) ...
[0];$left_arr=array();$right_arr=array();for($i=1;$i<count($arr);$i++){if($arr[$i]<=$key){$left_arr[]=$arr[$i];}else{$right_arr[]=$arr[$i];}}$left_arr=self::quickSort($left_arr);$right_arr=self::quickSort($right_arr);returnarray_merge($left_arr,array($key)...