First, move the array --- zero ※※※※※※

Given the nums an array, a write function to all moved to the end of the array 0, while maintaining the relative order of nonzero elements.

Example:

Input: [0,1,0,3,12]
Output: [1,3,12,0,0]
Description:

Must be operated on the original array, you can not copy additional array.
Minimize the number of operations.

Method: double pointer

i 0 points to the first;

j points to the first non-zero;

0 will move step by step to the right, one by one each treatment 0

. 1  class Solution {
 2  public :
 . 3      void moveZeroes (Vector < int > & the nums) {
 . 4          int I = 0 , J = 0 ;
 . 5          for (; J <nums.size (); J ++ )
 . 6          {
 . 7              the while (I < nums.size () && the nums [I] =! 0 ) I ++; // until the first 0, i is stopped 
. 8              the while (J <nums.size () && the nums [J] == 0 ) J ++; // is equal to 0, after the exchange is the position, so the time k ++, with continued to find a non-0 
. 9              IF (nums.size J == ()) BREAK ;
10              the else  IF (I> J) Continue ;
 . 11              the else { // do the exchange, they must be. 1 = J + I 
12 is                  int tmp = the nums [J];
 13 is                  the nums [J] = the nums [I];
 14                  the nums [ i] = tmp; // this time point i is non-zero after the switching, j points to the exchange over 0 
15                  i ++; // decrease a while, a bit save memory 
16              }
 17          }
 18      }

 

 

. 1  class Solution {
 2  public :
 . 3      void moveZeroes (Vector < int > & the nums) {
 . 4          int I = 0 ;
 . 5          for ( int J = 0 ; J <nums.size (); J ++) { // J encountered 0 when, at this time as the last round i ++, so this is actually encountered this time i a 0 
. 6              IF (the nums [J] =! 0 ) {
 . 7                  the nums [i] = the nums [J];
 . 8                  i ++ ;
 . 9              }
 10          }
 11          //This time point i is either 0 or i previously assigned non-zero, then after switching, i refers to 0; i refers to the short time that is 0 
12 is          the while (i < nums.size ()) {
 13 is                  the nums [I ++] = 0 ;
 14          }
 15          
16      }
 . 17 };