leetcode 88题

苏丛

1. #include <iostream>
   
2. #include <vector>
   
3. 
   
4. using namespace std;
   
5. 
   
6. class Solution {
   
7. public:
   
8.     void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
   
9.         int p1=m-1; // nums1未检查过的元素中最大的
   
10.         int p2=n-1; // nums2未检查过的元素中最大的
    
11. 
    
12.         // i为当前查看的元素
    
13.         for(int i=m+n-1; i>=0; i--){
    
14.             if(nums1[p1] > nums2[p2] or p2<0){
    
15.                 nums1[i] = nums1[p1--];
    
16.             }
    
17.             else{
    
18.                 nums1[i] = nums2[p2--];
    
19.             }
    
20.         }
    
21.     }
    
22. };
    
23. 
    
24. 
    
25. 
    
26. int main(){
    
27.     int arr1[] = {1,2,4,4,0,0,0};
    
28.     vector<int> vec1(arr1, arr1+sizeof(arr1)/sizeof(int));
    
29. 
    
30.     int arr2[] = {2,5,6};
    
31.     vector<int> vec2(arr2, arr2+sizeof(arr2)/sizeof(int));
    
32.    
    
33.     Solution().merge(vec1,4,vec2,3);
    
34. 
    
35.     for (int i = 0; i < vec1.size(); i++)
    
36.     {
    
37.         cout<<vec1[i]<<" ";
    
38.     }
    
39. }

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将两个数组合并, 在我的机器上跑没问题, 在leetcode上死活Runtime Error

题目大概是这样:
Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

https://leetcode.com/problems/merge-sorted-array/
求大佬指教

Seawolf

runtime error 也会有提示的，是哪个case 没有通过

苏丛

Seawolf 发表于 2019-10-6 07:29
runtime error 也会有提示的，是哪个case 没有通过

我那个example就通过不了
报的好像是内存的错误

1. Finished in N/A
   
2. =================================================================
   
3. ==29==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x6020000000ac at pc 0x00000040f85b bp 0x7ffffa5dcd10 sp 0x7ffffa5dcd08
   
4. READ of size 4 at 0x6020000000ac thread T0
   
5.     #1 0x7f94d173d2e0 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x202e0)
   
6. 0x6020000000ac is located 4 bytes to the left of 16-byte region [0x6020000000b0,0x6020000000c0)
   
7. allocated by thread T0 here:
   
8.     #0 0x7f94d3162ce0 in operator new(unsigned long) (/usr/local/lib64/libasan.so.5+0xe9ce0)
   
9.     #4 0x7f94d173d2e0 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x202e0)
   
10. Shadow bytes around the buggy address:
    
11.   0x0c047fff7fc0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
    
12.   0x0c047fff7fd0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
    
13.   0x0c047fff7fe0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
    
14.   0x0c047fff7ff0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
    
15.   0x0c047fff8000: fa fa fd fa fa fa fd fa fa fa fd fd fa fa fd fa
    
16. =>0x0c047fff8010: fa fa fd fa fa[fa]00 00 fa fa fa fa fa fa fa fa
    
17.   0x0c047fff8020: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
    
18.   0x0c047fff8030: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
    
19.   0x0c047fff8040: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
    
20.   0x0c047fff8050: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
    
21.   0x0c047fff8060: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
    
22. Shadow byte legend (one shadow byte represents 8 application bytes):
    
23.   Addressable:           00
    
24.   Partially addressable: 01 02 03 04 05 06 07
    
25.   Heap left redzone:       fa
    
26.   Freed heap region:       fd
    
27.   Stack left redzone:      f1
    
28.   Stack mid redzone:       f2
    
29.   Stack right redzone:     f3
    
30.   Stack after return:      f5
    
31.   Stack use after scope:   f8
    
32.   Global redzone:          f9
    
33.   Global init order:       f6
    
34.   Poisoned by user:        f7
    
35.   Container overflow:      fc
    
36.   Array cookie:            ac
    
37.   Intra object redzone:    bb
    
38.   ASan internal:           fe
    
39.   Left alloca redzone:     ca
    
40.   Right alloca redzone:    cb
    
41. ==29==ABORTING

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bin554385863

这个题是考察对容器操作的掌握
如果能熟练操作容器相关函数以及排序算法,这个问题一点都不难!
另外对容器增加元素时最好使用push_back()函数,而不是使用下标.

1. #include <iostream>
   
2. #include<algorithm>
   
3. #include <vector>
   
4. void merge(std::vector<int> &nums1, int m, std::vector<int> &nums2, int n)
   
5. {
   
6.     if( (m > nums1.size())||(n > nums2.size()))
   
7.     {
   
8.         exit(0);
   
9.     }
   
10.     nums1.erase(nums1.begin() + m, nums1.end());//擦除元素erase(begin, end)/erase(pos)函数
    
11.     nums2.erase(nums2.begin() + n, nums2.end());//擦除元素
    
12.     for(int i:nums2)
    
13.     {
    
14.         nums1.push_back(i);//合并容器
    
15.     }
    
16.     sort(nums1.begin(), nums1.end());//排序sort(begin, end, less<datetype>(降序)/greater<datetype>(升序))
    
17. }
    
18. int main(int argc, char const *argv[])
    
19. {
    
20.     std::vector<int> i = {0,1,2,3,9,10,11,0,0,0,0};
    
21.     std::vector<int> j = {4,5,6,7,23,54,0,0,0,0};
    
22.     merge(i,3,j,2);
    
23.     for(int k:i)
    
24.     {
    
25.         std::cout<<k;
    
26.     }
    
27.     return 0;
    
28. }

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---------------------------------------------------------------------------------------
E:\Users\86184\Documents\Code>c:\Users\86184\.vscode\extensions\ms-vscode.cpptools-0.25.1\debugAdapters\bin\WindowsDebugLauncher.exe --stdin=Microsoft-MIEngine-In-h1fynkur.lkj --stdout=Microsoft-MIEngine-Out-5krajmjw.k3n --stderr=Microsoft-MIEngine-Error-hbeytjdn.g0o --pid=Microsoft-MIEngine-Pid-twavuu23.nog "--dbgExe=E:\My Program\MinGW\bin\gdb.exe" --interpreter=mi
01245
E:\Users\86184\Documents\Code>
------------------------------------------------------------------------
执行结果：
通过
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执行用时 :
8 ms
, 在所有 C++ 提交中击败了
77.81%
的用户
内存消耗 :
8.6 MB
, 在所有 C++ 提交中击败了
89.69%
的用户

容器最基本的操作增察删改