原文:
给你一个长度为 n 的字符串 s(只有小写字母)。打印它必须获得的字符串的每个字符的位置，这样它将形成一个回文字符串。
示例:
input : c b b a a
output : 3 1 5 2 4
to make string palindrome 'c' must be at position 3,
'b' at 1 and 5, 'a' at 2 and 4.
input : a b c
output : not possible
any permutation of string cannot form palindrome.
其思想是创建一个向量数组(或动态大小数组)，存储每个字符的所有位置。在存储位置之后，我们检查奇数字符的计数是否超过 1。如果是，我们返回“不可能”。否则，我们首先打印数组的前半部分位置，然后打印奇数字符的一个位置(如果存在)，最后打印后半部分位置。
c
// cpp program to print original
// positions of characters in a
// string after rearranging and
// forming a palindrome
#include
using namespace std;
// maximum number of characters
const int max = 256;
void printpalindromepos(string &str)
{
// insert all positions of every
// character in the given string.
vector 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 1)
cout << "no palindrome";
/* print positions in first half
of palindrome */
for (int i=0; i 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
java 语言(一种计算机语言，尤用于创建网站)
// java program to print original
// positions of characters in a
// string after rearranging and
// forming a palindrome
import java.util.*;
class gfg
{
// maximum number of characters
static int max = 256;
static void printpalindromepos(string str)
{
// insert all positions of every
// character in the given string.
vector []pos = new vector[max];
for (int i = 0; i < max; i )
pos[i] = new vector();
int n = str.length();
for (int i = 0; i < n; i )
pos[str.charat(i)].add(i 1);
/* find the number of odd elements.
takes o(n) */
int oddcount = 0;
char oddchar = 0;
for (int i = 0; i < max; i )
{
if (pos[i].size() % 2 != 0)
{
oddcount ;
oddchar = (char) i;
}
}
/* a palindrome cannot contain more than 1
odd characters */
if (oddcount > 1)
system.out.print("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 )
system.out.print(pos[i].get(j) " ");
}
// consider one instance odd character
if (oddcount > 0)
{
int last = pos[oddchar].size() - 1;
system.out.print(pos[oddchar].get(last) " ");
pos[oddchar].remove(pos[oddchar].size() - 1);
}
/* 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 )
system.out.print(pos[i].get(j) " ");
}
}
// driver code
public static void main(string[] args)
{
string s = "geeksgk";
printpalindromepos(s);
}
}
// this code is contributed by 29ajaykumar
c
// c# program to print original
// positions of characters in a
// string after rearranging and
// forming a palindrome
using system;
using system.collections.generic;
class gfg
{
// maximum number of characters
static int max = 256;
static void printpalindromepos(string str)
{
// insert all positions of every
// character in the given string.
list []pos = new list[max];
for (int i = 0; i < max; i )
pos[i] = new list();
int n = str.length;
for (int i = 0; i < n; i )
pos[str[i]].add(i 1);
/* find the number of odd elements.
takes o(n) */
int oddcount = 0;
char oddchar = (char)0;
for (int i = 0; i < max; i )
{
if (pos[i].count % 2 != 0)
{
oddcount ;
oddchar = (char) i;
}
}
/* a palindrome cannot contain more than 1
odd characters */
if (oddcount > 1)
console.write("no palindrome");
/* print positions in first half
of palindrome */
for (int i = 0; i < max; i )
{
int mid = pos[i].count / 2;
for (int j = 0; j < mid; j )
console.write(pos[i][j] " ");
}
// consider one instance odd character
if (oddcount > 0)
{
int last = pos[oddchar].count - 1;
console.write(pos[oddchar][last] " ");
pos[oddchar].removeat(pos[oddchar].count - 1);
}
/* print positions in second half
of palindrome */
for (int i = max - 1; i >= 0; i--)
{
int count = pos[i].count;
for (int j = count / 2; j < count; j )
console.write(pos[i][j] " ");
}
}
// driver code
public static void main(string[] args)
{
string s = "geeksgk";
printpalindromepos(s);
}
}
// this code is contributed by 29ajaykumar
java 描述语言
output:
2 1 4 5 7 6 3
时间复杂度 : o ( n )
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