给定一个数字字符串,其中一些数字用“ $”代替,任务是通过用给定字符串的任何数字替换“ $”来生成所有可能的数字。
例子:
Input: str = “23$$”
Output:
2322
2323
2332
2333
Input: str = “$45”
Output:
445
545
方法:
- 找到字符串的所有组合通过与任何字符串的数字替换字符$,在本次检查,如果当前字符是一个数字,如果是,那么这个字符存储到一个数组预[]然后递归找到所有的组合否则,如果当前字符为“ $”,则将其替换为存储在数组中的数字,然后递归查找所有组合。
- 要查找所有可能的数字,请初始化存储所有可能数字的数组set [] ,要生成数字,请使用两个嵌套循环,其中外循环用于输入字符串,内循环用于存储所有数字可能组合的数组set [] 。初始化布尔标志来检查是否输入字符串的字符是在集已经存在[]或不如果输入字符串的字符是在集已经存在[]然后设定标志= FALSE否则,如果标志为真然后移动的当前字符输入要设置为set []的字符串,然后递归地找到输入字符串的所有组合,并将其存储在数组set []中。最后打印数组集中的每个数字[]
下面是上述方法的实现:
// C++ implementation of the approach
#include
#include
#include
#define MAX 20
#define DIGITS 10
// Array to store all the
// possible numbers
char set[DIGITS];
// Index to set[] element
int end;
// Function to find all the combinations
// of the string by replacing '$' with
// the other digits of the string
void combinations(char* num, char* pre, int curr, int lvl)
{
// Check if current length is less than
// the length of the input string
if (curr < strlen(num)) {
// If current charecter is a digit
// then store digit into pre[] and
// recursively find all the combinations
if (num[curr] >= '0' && num[curr] <= '9') {
pre[lvl] = num[curr];
combinations(num, pre, curr + 1, lvl + 1);
}
// If current charecter is a '$' then replace
// it with the other digits of the string and
// recursively find all the combinations
// Else go to the next charecter and
// recursively find all the combinations
else if (num[curr] == '$')
for (int i = 0; i < end; i++) {
pre[lvl] = set[i];
combinations(num, pre, curr + 1, lvl + 1);
}
else
combinations(num, pre, curr + 1, lvl);
}
// Print the array pre[]
else {
pre[lvl] = '\0';
printf("%s\n", pre);
}
}
// Function to find all the numbers formed
// from the input string by replacing '$' with
// all the digits of the input string
int findNumUtil(char num[])
{
// Array that stores the digits before
// the charecter $ in the input string
char pre[MAX];
end = 0;
// Traverse the input string and check if
// the current charecter is a digit
// if it is then set flag to true
for (int i = 0; i < strlen(num); i++)
if (num[i] >= '0' && num[i] <= '9') {
bool flag = true;
// Check if current character of the input
// string is already present in the array set[]
// then set flag to false
for (int j = 0; j < end; j++)
if (set[j] == num[i])
flag = false;
// Flag is true then store the charecter
// into set[] and recursively find all
// the combinations of numbers and store
// it in the set[] array
if (flag == true)
set[end++] = num[i];
}
combinations(num, pre, 0, 0);
return 0;
}
// Function to print all the combinations
// of the numbers by replacing '$' with
// the other digits of the input string
int findNum(char* num, int* result_count)
{
int i;
if (num[i]) {
result_count[i] = findNumUtil(num);
return (result_count[i]);
}
return 0;
}
// Driver code
int main()
{
char num[MAX] = "23$$";
int result_count[MAX];
findNum(num, result_count);
return 0;
}
输出:
2322
2323
2332
2333
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