leetcode 88题
#include <iostream>#include <vector>
using namespace std;
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int p1=m-1; // nums1未检查过的元素中最大的
int p2=n-1; // nums2未检查过的元素中最大的
// i为当前查看的元素
for(int i=m+n-1; i>=0; i--){
if(nums1 > nums2 or p2<0){
nums1 = nums1;
}
else{
nums1 = nums2;
}
}
}
};
int main(){
int arr1[] = {1,2,4,4,0,0,0};
vector<int> vec1(arr1, arr1+sizeof(arr1)/sizeof(int));
int arr2[] = {2,5,6};
vector<int> vec2(arr2, arr2+sizeof(arr2)/sizeof(int));
Solution().merge(vec1,4,vec2,3);
for (int i = 0; i < vec1.size(); i++)
{
cout<<vec1<<" ";
}
}
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 = , m = 3
nums2 = , n = 3
Output:
https://leetcode.com/problems/merge-sorted-array/
求大佬指教
runtime error 也会有提示的，是哪个case 没有通过 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
Finished in N/A
=================================================================
==29==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x6020000000ac at pc 0x00000040f85b bp 0x7ffffa5dcd10 sp 0x7ffffa5dcd08
READ of size 4 at 0x6020000000ac thread T0
#1 0x7f94d173d2e0 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x202e0)
0x6020000000ac is located 4 bytes to the left of 16-byte region [0x6020000000b0,0x6020000000c0)
allocated by thread T0 here:
#0 0x7f94d3162ce0 in operator new(unsigned long) (/usr/local/lib64/libasan.so.5+0xe9ce0)
#4 0x7f94d173d2e0 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x202e0)
Shadow bytes around the buggy address:
0x0c047fff7fc0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c047fff7fd0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c047fff7fe0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c047fff7ff0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c047fff8000: fa fa fd fa fa fa fd fa fa fa fd fd fa fa fd fa
=>0x0c047fff8010: fa fa fd fa fa00 00 fa fa fa fa fa fa fa fa
0x0c047fff8020: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c047fff8030: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c047fff8040: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c047fff8050: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c047fff8060: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
Addressable: 00
Partially addressable: 01 02 03 04 05 06 07
Heap left redzone: fa
Freed heap region: fd
Stack left redzone: f1
Stack mid redzone: f2
Stack right redzone: f3
Stack after return: f5
Stack use after scope: f8
Global redzone: f9
Global init order: f6
Poisoned by user: f7
Container overflow: fc
Array cookie: ac
Intra object redzone: bb
ASan internal: fe
Left alloca redzone: ca
Right alloca redzone: cb
==29==ABORTING 本帖最后由 bin554385863 于 2019-10-6 13:52 编辑
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.
#include <iostream>
#include<algorithm>
#include <vector>
void merge(std::vector<int> &nums1, int m, std::vector<int> &nums2, int n)
{
if( (m > nums1.size())||(n > nums2.size()))
{
exit(0);
}
nums1.erase(nums1.begin() + m, nums1.end());//擦除元素erase(begin, end)/erase(pos)函数
nums2.erase(nums2.begin() + n, nums2.end());//擦除元素
for(int i:nums2)
{
nums1.push_back(i);//合并容器
}
sort(nums1.begin(), nums1.end());//排序sort(begin, end, less<datetype>(降序)/greater<datetype>(升序))
}
int main(int argc, char const *argv[])
{
std::vector<int> i = {0,1,2,3,9,10,11,0,0,0,0};
std::vector<int> j = {4,5,6,7,23,54,0,0,0,0};
merge(i,3,j,2);
for(int k:i)
{
std::cout<<k;
}
return 0;
}
---------------------------------------------------------------------------------------
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
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