我的任务是输出所有十位数字,其中数字不重复 . 我首先使用的是这样的东西:
#include <cstdio>
#include <iostream>
#include <string>
#include <map>
#include <functional>
using namespace std;
void Task5() {
auto initialization = [](map<int, bool> *m, int count) {
for (int i = 0; i < 10; ++i)
m[i] = true;
};
/*For cut duplicate number in map*/
auto cutting = [](map<int, bool> *m, int count, int value) {
for (int i = 9; i > count; --i)
m[count][i][value] = false;
};
/*For create copy map*/
auto mould = [](map<int,bool> *m, map<int, bool> *m_copy, int count) -> map<int, bool>* {
if (m_copy == nullptr) {
map<int, bool> *m_copy = new map<int, bool>[10 - count];
for (int i = 9; i > count; --i)
for (int j = 0; j < 10; ++j)
m_copy[i][j] = m[i][j]; /*<= here throw exepition*/
return m_copy;
}
else {
for (int i = 9; i > count; --i)
for (int j = 0; j < 10; ++j)
m[i][j] = m_copy[i][j];
return m;
}
};
function<void(map<int, bool>*, int, int*)> recursive;
recursive = [mould, cutting, &recursive](map<int, bool> *m, int count = 1, int *result = nullptr) -> void {
if (count != 10) {
for (int i = 0; i < 10; ++i) {
static map<int, bool> *m_copy;
if (i == 0)
m_copy = mould(m, nullptr, 1);
else {
m = mould(m, m_copy, 1);
if (m[count][i])
result[count - 1] = i;
else
continue;
}
cutting(m, count, i);
recursive(m, ++count, result);
}
delete[] m_copy;
}
else {
for (int i = 0; i < 10; ++i)
cout << result[i];
cout << endl;
}
};
/*Create map
int is digit(can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
if digit is used bool will be false*/
map<int, bool> *m = new map<int, bool>[10];
for (int i = 0; i > 10; ++i)
initialization(m, i);
m[0][0] = false; //First number cant' be 0
int *result = new int[10];
recursive(m, 1, result);
delete[] m;
delete[] result;
}
int main(){
Task5();
return 0;
}
但它抛出了exdition std :: out_of_range . 现在我看了, Map [0]的大小为1,其他 Map ( Map [1], Map [2]等)的大小为0.为什么会这样?所以我期待论坛,找不到答案 . 所以我决定改写解决方案 . 写下这样的东西:
#include <cstdio>
#include <iostream>
#include <string>
#include <map>
#include <functional>
#include <vector>
using namespace std;
auto end_task = []() {
cout << endl << endl << endl;
};
void initialization(vector<bool> &vec) {
vec.reserve(10);
for (int i = 0; i < 10; ++i)
vec[i] = true;
}
void cutting(vector<bool> *vec, int count, int value) {
for (int i = 9; i > count; --i)
vec[i][value] = false;
}
vector<bool> *mould(vector<bool> *vec, vector<bool> *vec_copy, int count) {
if (vec_copy == nullptr) {
vector<bool> *vec_copy = new vector<bool>[10 - count];
for (int i = 9; i > count; --i)
for (int j = 0; j < 10; ++j)
vec_copy[i][j] = vec[i][j];
return vec_copy;
}
else {
for (int i = 9; i > count; --i)
for (int j = 0; j < 10; ++j)
vec[i][j] = vec_copy[i][j];
return vec;
}
}
void recursive(vector<bool> *vec, int count = 1, int *result = nullptr) {
if (count != 10) {
for (int i = 0; i < 10; ++i) {
static vector<bool> *vec_copy;
if (i == 0)
vec_copy = mould(vec, nullptr, 1);
else {
vec = mould(vec, vec_copy, 1);
if (vec[count][i])
result[count - 1] = i;
else
continue;
}
cutting(vec, count, i);
recursive(vec, ++count, result);
}
delete[] vec_copy;
}
else {
for (int i = 0; i < 10; ++i)
cout << result[i];
cout << endl;
}
}
void Task5() {
vector<bool> *vec = new vector<bool>[10];
for (int i = 0; i > 10; ++i)
initialization(vec[i]);
vec[0][0] = false;
int *result = new int[10];
recursive(vec, 1, result);
delete[] m;
delete[] result;
end_task();
}
int main(){
Task5();
return 0;
}
(没有lambda函数,因为我开始怀疑它们)但是这里的矢量大小是1和0.而且我有错误:矢量迭代器不是dereferencable . 为什么?我的错误在哪里?
2 回答
让我们仔细看看你得到异常的那一行(以及一些上下文):
第一行创建了一个名为
m_copy
的全新变量(由于它隐藏了同名的lambda参数而增加了混淆)并使其指向"array"的10 - count
元素 ."array"的顶部索引将是
10 - count - 1
,如果count == 0
,它将仅等于9
. 这意味着外部循环将在count > 0
的任何时候以无效和越界索引开始 .使用STL解决问题的简单方法是使用std::next_permutation:
Demo