用于查询给定字符串在恒定时间内的旋转和第 K 个字符的 C++ 程序
给定字符串str ,任务是对给定字符串执行以下类型的查询:
- (1, K):将字符串左旋转K个字符。
- (2, K):打印字符串的第 K个字符。
例子:
Input: str = “abcdefgh”, q[][] = {{1, 2}, {2, 2}, {1, 4}, {2, 7}}
Output:
d
e
Query 1: str = “cdefghab”
Query 2: 2nd character is d
Query 3: str = “ghabcdef”
Query 4: 7th character is e
Input: str = “abc”, q[][] = {{1, 2}, {2, 2}}
Output:
a
方法:这里的主要观察是字符串不需要在每个查询中旋转,而是我们可以创建一个指针ptr指向字符串的第一个字符,并且可以在每次旋转时更新它ptr = (ptr + K ) %N其中K是字符串需要旋转的整数, N是字符串的长度。现在对于第二种类型的每个查询,可以通过str[(ptr + K – 1) % N]找到第 K个字符。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
#define size 2
// Function to perform the required
// queries on the given string
void performQueries(string str, int n,
int queries[][size], int q)
{
// Pointer pointing to the current starting
// character of the string
int ptr = 0;
// For every query
for (int i = 0; i < q; i++) {
// If the query is to rotate the string
if (queries[i][0] == 1) {
// Update the pointer pointing to the
// starting character of the string
ptr = (ptr + queries[i][1]) % n;
}
else {
int k = queries[i][1];
// Index of the kth character in the
// current rotation of the string
int index = (ptr + k - 1) % n;
// Print the kth character
cout << str[index] << "
";
}
}
}
// Driver code
int main()
{
string str = "abcdefgh";
int n = str.length();
int queries[][size] = { { 1, 2 }, { 2, 2 },
{ 1, 4 }, { 2, 7 } };
int q = sizeof(queries) / sizeof(queries[0]);
performQueries(str, n, queries, q);
return 0;
}
输出:
d
e
有关详细信息,请参阅关于给定字符串的旋转和第 K 个字符的查询的完整文章!