merge.printAll(array); }publicvoidMerge(int[] array,intlow,intmid,inthigh) {inti = low;//i是第一段序列的下标intj = mid + 1;//j是第二段序列的下标intk = 0;//k是临时存放合并序列的下标int[] array2 =newint[high - low + 1];//array2是临时合并序列//扫描第一段和第二段序列,直到...
voidmerge(intA[],intm,intB[],intn){intC[100000];inti=0,j=0;while(i < m && j <n){if(A[i] <=B[j]){ C[i+j] =A[i]; i++; }else{ C[i+j] =B[j]; j++; } }while(i <m){ C[i+j] =A[i]; i++; }while(j <n){ C[i+j] =B[j]; j++; }for(i =0; i...
print(id(nums1)) nums1 = sorted(nums1) print(id(nums1))#此时nums1的地址被sorted改变,所以nums1是新地址,不符合题中不能用新空间 #print(nums1) nums1 = nums1[(len1-m-n):] print(id(nums1))#肯定首地址都发生变化了 return nums1 #print(nums1) 1. 2. 3. 4. 5. 6. 7. 8. 9....
leetcode 88. Merge Sorted Array 合并到一个新的数组,直接比较就好了,这个题目是将nums1、nums2合并到nums1,nums1有许多多余的空间 如果按照合并到一个新的数组从小比到大的方式进行比较,就需要每次挪动nums1的数组。 本题可以采用从大到小的比较方式,这样就不用每次挪动数组。 同时注意,m和n都是可以为0的,...
88. Merge Sorted Array Given two sorted integer arraysnums1andnums2, mergenums2intonums1as one sorted array. Note: The number of elements initialized innums1andnums2aremandnrespectively. You may assume thatnums1has enough space (size that is greater or equal tom+n) to hold additional elemen...
7. Merge Two Sorted Arrays (Descending)Write a program in C to merge two arrays of the same size sorted in descending order.This task requires writing a C program to merge two arrays of the same size and sort the resulting array in descending order. The program will take inputs for ...
merge()是C++标准库的函数，主要实现函数的排序和合并，不仅仅是合并，具体要求参照标准库。include"stdafx.h"include<iostream> include<algorithm> include<array> include<list> usingnamespacestd;boolcomp(constinti,constintj){ returni>j;} intmain(void){ /*自定义谓词*/ std::array<int,4>...
//本程序实现了既可以保留merging过程中的temp⽂件，也提供的删除temp⽂件，只保留sorted 和unsorted 的⽂件的⽅式，具体调整在函数clear_external_memory()中，//可以注释掉，也可以让其运⾏ /*Group 1. An attempt to sort a large file of integers, breaking the large file into arrays of 1000 ...
时间复杂度 还是这张图，可以看出，一共有logn层，每一层合并的复杂度为O(n) ，也就是 T(n) = 2T(n/2) + O(n) = 2(2T(n/4) + O(n/2)) + O(n) = ... 假设n = 1时，T(1) = c，则T(n) = 2T(n/2) + cn = 2(2T(n/4) + cn/2)) + cn = ... =cnlog...
The source code was compiled using g++ -O3 -w -fpermissive bench.c. It measures the performance on random data with array sizes ranging from 10 to 10,000,000. It's generated by running the benchmark using 10000000 0 0 as the argument. The benchmark is weighted, meaning the number of ...