回文是一個與其反轉相等的字串。給定一個字串，我們需要找到使該字串成為回文所需的最小插入任意字元的數量。我們將看到三種方法：首先是遞歸方法，然後我們將記憶化這個解決方案，最後，我們將實現動態規劃方法。
#include <stdio.h> // library for input and output
#include <limits.h> // library to get the integer limits
#include <string.h> // library for strings
// function to find the minimum of two number
// as it is not present in the c language
int findMin(int a, int b){
if(a < b){
return a;
} else{
return b;
}
}
// creating the function to find the required answer we will make recursive calls to it
int findAns(char str[], int start, int end){
// base condition
if (start > end){
return INT_MAX;
}
else if(start == end){
return 0;
}
else if (start == end - 1){
if(str[start] == str[end]){
return 0;
}
else return 1;
}
// check if both start and end characters are the same make callson the basis of that
if(str[start] == str[end]){
return findAns(str,start+1, end-1);
} else{
return 1+ findMin(findAns(str,start,end-1), findAns(str,start+1,end));
}
}
// main function
int main(){
char str[] = "thisisthestring"; // given string
printf("The minimum number of insertions required to form the palindrome is: %d", findAns(str,0,strlen(str)-1));
return 0;
}
The minimum number of insertions required to form the palindrome is: 8
以上程式碼的時間複雜度為O(2^N)，因為我們對每個插入都進行選擇，其中N是給定字串的大小。
上述程式碼的空間複雜度為O(N)，即在遞迴呼叫中使用。
#include <stdio.h> // library for input and output
#include <limits.h> // library to get the integer limits
#include <string.h> // library for strings
int memo[1005][1005]; // array to store the recursion results
// function to find the minimum of two number
// as it is not present in the c language
int findMin(int a, int b){
if(a < b){
return a;
} else{
return b;
}
}
// creating the function to find the required answer we will make recursive calls to it
int findAns(char str[], int start, int end){
// base condition
if (start > end){
return INT_MAX;
}
else if(start == end){
return 0;
}
else if (start == end - 1){
if(str[start] == str[end]){
return 0;
}
else return 1;
}
// if already have the result
if(memo[start][end] != -1){
return memo[start][end];
}
// check if both start and end characters are same make calls on basis of that
if(str[start] == str[end]){
memo[start][end] = findAns(str,start+1, end-1);
} else{
memo[start][end] = 1+ findMin(findAns(str,start,end-1), findAns(str,start+1,end));
}
return memo[start][end];
}
int main(){
char str[] = "thisisthestring"; // given string
//Initializing the memo array
memset(memo,-1,sizeof(memo));
printf("The minimum number of insertions required to form the palindrome is: %d", findAns(str,0,strlen(str)-1));
return 0;
}
The minimum number of insertions required to form the palindrome is: 8
上述程式碼的時間複雜度為O(N^2)，因為我們儲存了已經計算過的結果。
上述程式碼的空間複雜度為O(N^2)，因為我們在這裡使用了額外的空間。
#include <stdio.h> // library for input and output
#include <limits.h> // library to get the integer limits
#include <string.h> // library for strings
// function to find the minimum of two number
// as it is not present in the c language
int findMin(int a, int b){
if(a < b){
return a;
} else{
return b;
}
}
// creating a function to find the required answer
int findAns(char str[], int len){
// creating the table and initialzing it
int memo[1005][1005];
memset(memo,0,sizeof(memo));
// filling the table by traversing over the string
for (int i = 1; i < len; i++){
for (int start= 0, end = i; end < len; start++, end++){
if(str[start] == str[end]){
memo[start][end] = memo[start+1][end-1];
} else{
memo[start][end] = 1 + findMin(memo[start][end-1], memo[start+1][end]);
}
}
}
// return the minimum numbers of interstion required for the complete string
return memo[0][len-1];
}
int main(){
char str[] = "thisisthestring"; // given string
// calling to the function
printf("The minimum number of insertions required to form the palindrome is: %d", findAns(str, strlen(str)));
return 0;
}
The minimum number of insertions required to form the palindrome is: 8
上述程式碼的時間複雜度為O(N^2)，因為我們在這裡使用了嵌套的for迴圈。
上述程式碼的空間複雜度為O(N^2)，因為我們在這裡使用了額外的空間。
在本教程中，我們實作了三種方法來找到使給定字串成為回文所需的最小插入次數。我們實作了遞歸方法，然後進行了記憶化處理。最後，我們實現了表格法或動態規劃法。
以上是C程式找出形成回文的最小插入次數的詳細內容。更多資訊請關注PHP中文網其他相關文章！
