[Leetcode] Move Zeroes, leetcodezeroes

[Leetcode] Move Zeroes, leetcodezeroes
Move Zeroes questions:

Given an arraynums, Write a function to move all0'S to the end of it while maintaining the relative order of the non-zero elements.

For example, givennums = [0, 1, 0, 3, 12], After calling your function,numsShocould be[1, 3, 12, 0, 0].

Note:
The idea at the beginning was to traverse the array, move it back when it encounters 0, and keep moving the last step. The Code is as follows:

public class Solution {    public static void moveZeroes(int[] nums) {        for (int i = 0; i < nums.length; i++) {            if (nums[i] == 0 && i != (nums.length - 1)) {                for (int j = i; j < nums.length - 1; j++) {                    int temp = nums[j];                    nums[j] = nums[j + 1];                    nums[j + 1] = temp;                }            }        }    }}

However, the 001 test case is incorrect.

Well, I found the problem, because the first number is 0, and the second 0 cannot be moved to the end.

In another way, instead of looking for 0 through traversal, we traverse to find non-0. If it is not 0 and is not in the first place, we will move forward until it is not 0.

public class Solution {    public static void moveZeroes(int[] nums) {        for (int i = 0; i < nums.length; i++) {            if (nums[i] != 0 && i != 0) {                for (int j = i; j > 0 && nums[j - 1] == 0; j--) {                    int temp = nums[j - 1];                    nums[j - 1] = nums[j];                    nums[j] = temp;                }            }        }    }}

This solves the problem.

Other methods

public static void moveZeroes(int[] nums) {        int i = 0, j = 0;        for (i = 0; i < nums.length; i++) {            if (nums[i] != 0) {                if (j != i) {                    nums[j] = nums[i];                    nums[j] = 0;                }                j++;            }        }    }

In this method, I traverses the array. Only when I points to a non-zero and ji is different, j is moved back and the number at j is changed to the number at I, and set the number at I to 0 and ij to synchronize and move backward. When I points to 0, only I moves backward and j still points to 0.

There is also a simpler method

public void moveZeroes(int[] nums) {    if (nums == null || nums.length == 0) return;            int insertPos = 0;    for (int num: nums) {        if (num != 0) nums[insertPos++] = num;    }            while (insertPos < nums.length) {        nums[insertPos++] = 0;    }}