Question

I am given a string S and I have to make it a palindrome by re-arranging the characters of the given string.

If the given string can not be converted into a palindrome by re-arranging characters, print false and if it is possible to make it a palindrome, print true

My code:

       String a=new String(br.readLine());      //given string
   int n= a.length();   
                    int j=0,k=n-1,count=0;
        boolean flag=false;
        for(int i=0;i<n;i++)
        {
            if(a.charAt(i)=='*')
                continue;        //for skipping already shifted  chars
            int ix = a.indexOf(a.charAt(i), i+1);
            if(ix >= 0)
            {                   a=a.substring(0,i+1)+a.substring(i+1, ix) + "*" + a.substring(ix+1);
            }
            else
            {
                count++;        //number of unique chars which can only be 1 or 0
                if(count<=1 && n%2==1)
                {

                    a=a.replaceFirst(a.substring(i,i+1),"*");                  //giving middle position to the only unique char at center and replacing it with *
                }
                else
                {
                    System.out.println("false");    //if more than one unique char, palindrome not possible
                    flag=true; // shows not possible
                    break;
                }
            }

        }
        if(!flag) // if possible
        {                   
            System.out.println("true");
        }

Answer 1

一个明显的优化将取代：

        if(a.substring(i+1).contains(a.substring(i,i+1)))
        {
            ans[j++]=i;                           //storing new positions in ans array
            ans[k--]=a.substring(i+1).indexOf(a.charAt(i))+1+i;
            a=a.substring(0,i+1)+a.substring(i+1).replaceFirst(a.substring(i,i+1),"*");                 //replacing the shifted char with *
        }

使用：

        int ix = a.indexOf(a.charAt(i), i+1);
        if(ix >= 0)
        {
            ans[j++]=i;                           //storing new positions in ans array
            ans[k--]=ix;
            a=a.substring(0,i+1)+a.substring(i+1, ix) + "*" + a.substring(ix+1);
        }

更新

我想知道以下代码是否会更快。数组上没有indexOf，所以我不得不做一个循环，但是没有字符串操作：

    char[] c = a.toCharArray();
    int n= c.length;  
    int ans[]=new int[n];  // for storing new positions after shifting
    int j=0,k=n-1,count=0;
    boolean flag=false;
    for(int i=0; i < n; i++)
    {
        char ch = c[i];
        if(ch=='*')
            continue;        //for skipping already shifted  chars
        int ix = i;
        do {
            ++ix;
        } while (ix < n && c[ix] != ch);
        if(ix < n)
        {
            ans[j++]=i;                          
            ans[k--]=ix;
            c[ix] = '*';
        }
        else
        {
            count++;        //number of unique chars which can only be 1 or 0
            if(count<=1 && n%2==1)
            {
                ans[(int)n/2]=i;
                c[i] = '*';
            }
            else
            {
                System.out.println("-1");   //if more than one unique char, palindrome not possible
                flag=true; // shows not possible
                break;
            }
        }

    }

更新2

当j到达n / 2时，您也可以停止：

    char[] c = a.toCharArray();
    int n= c.length;  
    int ans[]=new int[n];  // for storing new positions after shifting
    int j=0,k=n-1,count=0,half=n/2;
    boolean flag=false;
    for(int i=0; i < n; i++)
    {
        char ch = c[i];
        if(ch=='*')
            continue;        //for skipping already shifted  chars
        int ix = i;
        do {
            ++ix;
        } while (ix < n && c[ix] != ch);
        if(ix < n)
        {
            ans[j++]=i;                          
            ans[k--]=ix;
            c[ix] = '*';
            if (j > half) {
                break;
            }
        }
        else
        {
            count++;        //number of unique chars which can only be 1 or 0
            if(count<=1 && n%2==1)
            {
                ans[half]=i;
                c[i] = '*';
            }
            else
            {
                System.out.println("-1");   //if more than one unique char, palindrome not possible
                flag=true; // shows not possible
                break;
            }
        }

    }

making a string samefrom both ends by shifting characters

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