移动零
问题描述：
给定一个数组
nums，编写一个函数将所有0移动到数组的末尾，同时保持非零元素的相对顺序。请注意 ，必须在不复制数组的情况下原地对数组进行操作。示例 1:
输入: nums = [0,1,0,3,12] 输出: [1,3,12,0,0]示例 2:
输入: nums = [0] 输出: [0]
//解法一：
class Solution {public int[] moveZeroes(int[] nums){int n = nums.length;int j =0; //指针j为指向已经处理好的序列尾部//指针i指向未处理好的序列头部for (int i = 0;i < n; i++){if(nums[i] != 0){int temp = nums[i];nums[i] = nums[j];nums[j] = temp;j++;}}return nums;}
}//解法二：
class Solution {public int[] moveZeroes(int[] nums){int n = nums.length;int j =0;for (int i = 0;i < n; i++){if(nums[i] != 0){nums[j] = nums[i];//指针j指向为0的数，直接将其变为非0的数j++;}}//第一次循环将前面所有不为0的数全部移动到前面，且指针j最后指向的是已经处理好的序列尾部//所以指针i和指针j之间的数为0的个数for (int i = j; i<n; i++){nums[i] = 0;}return nums;}
}//带有输入输出：
//解法一：
public class hot4_moveZeroes {public int[] moveZeroes(int[] nums){int n = nums.length;int j =0;for (int i = 0;i < n; i++){if(nums[i] != 0){int temp = nums[i];nums[i] = nums[j];nums[j] = temp;j++;}}return nums;}public static void main(String[] args) {hot4_moveZeroes hot4MoveZeroes = new hot4_moveZeroes();int[] nums = {0,1,0,3,12};System.out.println("输入： = " + Arrays.toString(nums));int[] result = hot4MoveZeroes.moveZeroes(nums);System.out.println("输出：" + Arrays.toString(result));}
}//解法二：
public class hot4_moveZeroes {public int[] moveZeroes(int[] nums){int n = nums.length;int j =0;for (int i = 0;i < n; i++){if(nums[i] != 0){nums[j] = nums[i];j++;}}for (int i = j; i<n; i++){nums[i] = 0;}return nums;}public static void main(String[] args) {hot4_moveZeroes hot4MoveZeroes = new hot4_moveZeroes();int[] nums = {0,1,0,3,12};System.out.println("输入： = " + Arrays.toString(nums));int[] result = hot4MoveZeroes.moveZeroes(nums);System.out.println("输出：" + Arrays.toString(result));}
}
知识点总结：
- 方法二的思想
