如果可能，重新排列字符以形成回文
给定一个字符串，将字符串转换为回文而不做任何修改，如添加字符、删除字符、替换字符等。
例子：
Input : "mdaam"
Output : "madam" or "amdma"
Input : "abb"
Output : "bab"
Input : "geeksforgeeks"
Output : "No Palindrome" 1.计算所有字符的出现次数。
2. 计算奇数。如果这个计数大于 1 或等于 1 并且字符串的长度是偶数，那么显然回文不能从给定的字符串形成。
3.初始化两个空字符串firstHalf和secondHalf。
4. 遍历地图。对于 count 为 count 的每个字符，将 count/2 个字符附加到 firstHalf 的末尾和 secondHalf 的开头。
5.最后通过附加firstHalf和secondHalf返回结果
C++
// C++ program to rearrange a string to
// make palindrome.
#include
using namespace std;
string getPalindrome(string str)
{
// Store counts of characters
unordered_map hmap;
for (int i = 0; i < str.length(); i++)
hmap[str[i]]++;
/* find the number of odd elements.
Takes O(n) */
int oddCount = 0;
char oddChar;
for (auto x : hmap) {
if (x.second % 2 != 0) {
oddCount++;
oddChar = x.first;
}
}
/* odd_cnt = 1 only if the length of
str is odd */
if (oddCount > 1
|| oddCount == 1 && str.length() % 2 == 0)
return "NO PALINDROME";
/* Generate first halh of palindrome */
string firstHalf = "", secondHalf = "";
for (auto x : hmap) {
// Build a string of floor(count/2)
// occurrences of current character
string s(x.second / 2, x.first);
// Attach the built string to end of
// and begin of second half
firstHalf = firstHalf + s;
secondHalf = s + secondHalf;
}
// Insert odd character if there
// is any
return (oddCount == 1)
? (firstHalf + oddChar + secondHalf)
: (firstHalf + secondHalf);
}
int main()
{
string s = "mdaam";
cout << getPalindrome(s);
return 0;
} Java
// Java program to rearrange a string to
// make palindrome
import java.util.HashMap;
import java.util.Map.Entry;
class GFG{
public static String getPalindrome(String str)
{
// Store counts of characters
HashMap counting = new HashMap<>();
for(char ch : str.toCharArray())
{
if (counting.containsKey(ch))
{
counting.put(ch, counting.get(ch) + 1);
}
else
{
counting.put(ch, 1);
}
}
/* Find the number of odd elements.
Takes O(n) */
int oddCount = 0;
char oddChar = 0;
for(Entry itr : counting.entrySet())
{
if (itr.getValue() % 2 != 0)
{
oddCount++;
oddChar = itr.getKey();
}
}
/* odd_cnt = 1 only if the length of
str is odd */
if (oddCount > 1 || oddCount == 1 &&
str.length() % 2 == 0)
{
return "NO PALINDROME";
}
/* Generate first halh of palindrome */
String firstHalf = "", lastHalf = "";
for(Entry itr : counting.entrySet())
{
// Build a string of floor(count/2)
// occurrences of current character
String ss = "";
for(int i = 0; i < itr.getValue() / 2; i++)
{
ss += itr.getKey();
}
// Attach the built string to end of
// and begin of second half
firstHalf = firstHalf + ss;
lastHalf = ss + lastHalf;
}
// Insert odd character if there
// is any
return (oddCount == 1) ?
(firstHalf + oddChar + lastHalf) :
(firstHalf + lastHalf);
}
// Driver code
public static void main(String[] args)
{
String str = "mdaam";
System.out.println(getPalindrome(str));
}
}
// This code is contributed by Satyam Singh Python3
# Python3 program to rearrange a string to
# make palindrome.
from collections import defaultdict
def getPalindrome(st):
# Store counts of characters
hmap = defaultdict(int)
for i in range(len(st)):
hmap[st[i]] += 1
# Find the number of odd elements.
# Takes O(n)
oddCount = 0
for x in hmap:
if (hmap[x] % 2 != 0):
oddCount += 1
oddChar = x
# odd_cnt = 1 only if the length of
# str is odd
if (oddCount > 1 or oddCount == 1 and
len(st) % 2 == 0):
return "NO PALINDROME"
# Generate first halh of palindrome
firstHalf = ""
secondHalf = ""
for x in sorted(hmap.keys()):
# Build a string of floor(count/2)
# occurrences of current character
s = (hmap[x] // 2) * x
# Attach the built string to end of
# and begin of second half
firstHalf = firstHalf + s
secondHalf = s + secondHalf
# Insert odd character if there
# is any
if (oddCount == 1):
return (firstHalf + oddChar + secondHalf)
else:
return (firstHalf + secondHalf)
# Driver code
if __name__ == "__main__":
s = "mdaam"
print(getPalindrome(s))
# This code is contributed by ukaspJavascript
输出
amdma