leecode27 : https://leetcode.cn/problems/remove-element/submissions/521113648/

暴力解法

循环匹配，每次匹配到就将数组匹配到的元素的后面元素向前移动一位
【注意】 注意最后一位元素，避免数组越界

public class RemoveElement {public int removeElement(int[] nums, int val) {int count = 0;for (int i = 0; i < nums.length  - count; i++) {// 处理最后一个临界元素if(i == nums.length && nums[i] == val ) {count++;nums[i] = 0;return  count;}if (nums[i] == val) {count++;for (int j = i + 1; j < nums.length; j++) {nums[j - 1] = nums[j];}i = i - 1;}}return nums.length - count;}public static void main(String[] args) {RemoveElement r = new RemoveElement();int[] nums = {0, 1, 2, 2, 3, 0, 4, 2};int count = r.removeElement(nums, 2);System.out.println(count);for (int i = 0; i < count; i++) {System.out.print(nums[i] + " ");}}
}

双指针思路

package leecode.num27;// nums = [3,2,2,3], val = 3public class RemoveElement {// 双指针解法public int removeElement1(int[] nums, int val) {int fast = 0, slow = 0;for (fast = 0; fast < nums.length; fast++) {if (nums[fast] != val) {nums[slow] = nums[fast];slow++;}}return slow;}public static void main(String[] args) {RemoveElement r = new RemoveElement();int[] nums = {0, 1, 2, 2, 3, 0, 4, 2};int count = r.removeElement1(nums, 2);System.out.println(count);for (int i = 0; i < count; i++) {System.out.print(nums[i] + " ");}}
}