next-permutation

packagecom.ytx.array;

/**

 *  题目 ：   next-permutation

 * 

 *  描述：    Implement next permutation, which rearranges numbers into the lexicographically

 *           next greater permutation of numbers.

                    If such arrangement is not possible, it must rearrange it as the lowest possible

                    order (ie, sorted in ascending order).

                    The replacement must be in-place, do not allocate extra memory.

                    Here are some examples. Inputs are in the left-hand column and

                    its corresponding outputs are in the right-hand column.

                           1,2,3→1,3,2

                           3,2,1→1,2,3

                           1,1,5→1,5,1

                           

                           

 *@authoryuantian xin

 *

 *     题意就是说按照字典序(升序）生成当前排列的下一个排列，如果输入序列已经是降序排列，说明这个是最后一个排列，直接反序输出即可。

 *  1234->1243->1324->1342->1423->1432->4123->...->4321->1234;

 *

 */

publicclass Next_permutation {

       

       

       publicvoid nextPermutation(int[]num) {

             intlen = num.length;

       //从后往前找到第一个升序位置

             intpos = -1;

             for(inti = len - 1; i > 0; i--) {

                    if(num[i] > num[i-1]) {

                           pos= i - 1;

                           break;

                    }

             }

             

             //如果不存在升序，说明这个序列是最后一个排列，直接逆序这个序列并return。

             if(pos< 0) {

                    reverse(num, 0,len-1);

                    return;

             }

             

             //如果找到了升序位置，那么从后往前找，找到pos之后第一个比num[pos]大的位置，并且交换这两个位置上的数。

             for(inti = len-1;i > pos;i--) {

                    if(num[i] > num[pos]) {

                           inttemp = num[i];

                           num[i] = num[pos];

                           num[pos] = temp;

                           break;

                    }

             }

             //逆序pos位置之后的数字

             reverse(num,pos + 1, len-1);

             

    }  

   

       publicvoid reverse(int[]num,int begin, int end) {

             

             inttemp;

             

             while(begin< end) {

                    temp= num[begin];

                    num[begin] = num[end];

                    num[end] = temp;

                    begin++;

                    end--;

             }

             

       }

       publicstatic void main(String[]args) {

             /*int []num = {1,2,3,6,5,4};*/

             int[] num = {1,2,3,4,6,5};

             newNext_permutation().nextPermutation(num);

             for(inti = 0; i < num.length;i++) {

                    System.out.print(num[i] + " ");

             }

       }

}