【算法-数组】移除元素

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 = 3

public 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] + " ");
        }
    }
}