C++ qsort() 函数调用时实参与形参不兼容的问题解决

《剑指OFFER》刷题笔记 —— 扑克牌顺子

LL今天心情特别好,因为他去买了一副扑克牌,发现里面居然有2个大王,2个小王(一副牌原本是54张^_^)...他随机从中抽出了5张牌,想测测自己的手气,看看能不能抽到顺子,如果抽到的话,他决定去买体育彩票,嘿嘿！！“红心A,黑桃3,小王,大王,方片5”,“Oh My God!”不是顺子.....LL不高兴了,他想了想,决定大\小 王可以看成任何数字,并且A看作1,J为11,Q为12,K为13。上面的5张牌就可以变成“1,2,3,4,5”(大小王分别看作2和4),“So Lucky!”。LL决定去买体育彩票啦。 现在,要求你使用这幅牌模拟上面的过程,然后告诉我们LL的运气如何， 如果牌能组成顺子就输出true，否则就输出false。为了方便起见,你可以认为大小王是0。

ERROR:  VS 的错误提示 —— int (Solution::*) (const void* a, const void* b) 类型的实参与 int (_cdecl::*) (const void* , const void* ) 类型的形参不兼容****

1class Solution {
2public: 
3　　bool IsContinuous(vector<int> numbers) {
4        if (numbers.size() != 5)
5            return false;
6        //qsort(&numbers, numbers.size(), sizeof(int), compare);
7        qsort(&numbers[0], numbers.size(), sizeof(int), compare);
8        //sort(numbers.begin(), numbers.end());
9        
10        int numZeros = 0;
11        int diff = 0;
12        for (int i = 0; i < numbers.size() - 1; ++i) {
13            if (numbers[i] == 0)
14                ++ numZeros;
15            else {
16                if (numbers[i] == numbers[i + 1])
17                    return false;
18                diff += numbers[i + 1] - numbers[i] - 1;
19            }
20        }
21        return (numZeros < diff) ? false : true;
22    }
23　　
24　 int compare(const void *a, const void *b) {
25　　  return (*(int*)a - *(int*)b);
26　 }
27
28};

CORRECT:

1int compare(const void *a, const void *b) {
2　　return (*(int*)a - *(int*)b);
3}
4
5class Solution {
6public: 
7　　bool IsContinuous(vector<int> numbers) {
8        if (numbers.size() != 5)
9            return false;
10        //qsort(&numbers, numbers.size(), sizeof(int), compare);
11        qsort(&numbers[0], numbers.size(), sizeof(int), compare);
12        //sort(numbers.begin(), numbers.end());
13        
14        int numZeros = 0;
15        int diff = 0;
16        for (int i = 0; i < numbers.size() - 1; ++i) {
17            if (numbers[i] == 0)
18                ++ numZeros;
19            else {
20                if (numbers[i] == numbers[i + 1])
21                    return false;
22                diff += numbers[i + 1] - numbers[i] - 1;
23            }
24        }
25        return (numZeros < diff) ? false : true;
26    }
27　　
28};

注意：

C++ 排序请使用 sort()，不要使用 C 语言中的 qsort()；

转一下 https://stackoverflow.com/questions/12308243/trying-to-use-qsort-with-vector 的大佬评论：

First of all, DON'T.

If you just want to muck about, you can replace iterators with actual pointers:

qsort(&numbers[0], numbers.size(), sizeof(int), compvar);

Apart from not doing all the work std::sort does, there is one unexpected thing about qsort. It is slower.

1. sort (myvector1.begin(), myvector1.end());

2. sort (myvector2.begin(), myvector2.end(), myfunction);

3. sort (myvector3.begin(), myvector3.end(), myobject);

4. qsort(&myvector4[0], myvector4.size(), sizeof(int), cmyfunction);

4 is the slowest, followed by 2 (function pointer passed to std::sort). 1 and 3 (default and functor) are the fastest (compiled with gnu's g++ with -O3 flag).