为您提供了一个长度为 n 的字符串 str。打印字符串中每个元素的位置,以便它可以形成回文,否则在屏幕上打印消息“No palindrome”。
什么是回文?
Palindrome 是一个单词,从反向或向后读取的字符序列与从正向读取的字符序列相同,例如 MADAM、racecar。
要查找序列或单词是回文,我们通常将单词的反向存储在单独的字符串中并比较两者,如果它们相同,则给定的单词或序列是回文。但是在这个问题中,我们必须打印排列以形成回文中的单词或序列。
就像,有一个字符串 str = “tinni” 那么它可以是 intni 或 nitin 所以我们必须返回作为从 1 开始的索引和结果的任何一个排列顺序可以是 2 3 1 4 5 或 3 2 1 5 4 两者中的一个。
上述问题需要像下面给出的示例那样的解决方案-
示例
'Input: string str = “baa”
Output: 2 1 3
Input: string str = “tinni”
Output: 2 3 1 4 5
算法
'void printPalindromePos(string &str)
START
STEP 1: DECLARE vector<int> pos[MAX]
STEP 2: DECLARE AND ASSIGN n WITH LENGTH OF str
STEP 3: LOOP FOR i = 0 AND i < n AND i++
pos[str[i]].push_back(i+1)
END LOOP
STEP 4: SET oddCount = 0
STEP 5: DECLARE oddChar
STEP 6: LOOP FOR i=0 AND i<MAX AND i++
IF pos[i].size() % 2 != 0 THEN,
INCREMENT oddCount BY 1
SET oddChar AS i
END IF
END FOR
STEP 7: IF oddCount > 1 THEN,
PRINT "NO PALINDROME"
STEP 8: LOOP FOR i=0 AND i<MAX AND i++
DECRLARE mid = pos[i].size()/2
LOOP FOR j=0 AND j<mid AND j++
PRINT pos[i][j]
END LOOP
END LOOP
STEP 9: IF oddCount > 0 THEN,
DECLARE AND SET last = pos[oddChar].size() - 1
PRINT pos[oddChar][last]
SET pos[oddChar].pop_back();
END IF
STEP 10: LOOP FOR i=MAX-1 AND i>=0 AND i--
DECLARE AND SET count = pos[i].size()
LOOP FOR j=count/2 AND j<count AND j++
PRINT pos[i][j]
STOP
示例
'#include <bits/stdc++.h>
using namespace std;
// Giving the maximum characters
const int MAX = 256;
void printPalindromePos(string &str){
//Inserting all positions of characters in the given string.
vector<int> pos[MAX];
int n = str.length();
for (int i = 0; i < n; i++)
pos[str[i]].push_back(i+1);
/* find the number of odd elements.Takes O(n) */
int oddCount = 0;
char oddChar;
for (int i=0; i<MAX; i++) {
if (pos[i].size() % 2 != 0) {
oddCount++;
oddChar = i;
}
}
/* Palindrome can't contain more than 1 odd characters */
if (oddCount > 1)
cout << "NO PALINDROME";
/* Print positions in first half of palindrome */
for (int i=0; i<MAX; i++){
int mid = pos[i].size()/2;
for (int j=0; j<mid; j++)
cout << pos[i][j] << " ";
}
// Consider one instance odd character
if (oddCount > 0){
int last = pos[oddChar].size() - 1;
cout << pos[oddChar][last] << " ";
pos[oddChar].pop_back();
}
/* Print positions in second half of palindrome */
for (int i=MAX-1; i>=0; i--){
int count = pos[i].size();
for (int j=count/2; j<count; j++)
cout << pos[i][j] << " ";
}
}
int main(){
string s = "tinni";
printPalindromePos(s);
return 0;
}
输出
如果我们运行上面的程序,那么它将生成以下输出 -
'2 3 1 4 5