LeetCode 27. Remove Element

最新推荐文章于 2024-02-22 13:25:37 发布

Starry_1

最新推荐文章于 2024-02-22 13:25:37 发布

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分类专栏： LeetCode刷题文章标签： Java LeetCode 刷题

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本文介绍了一种在O(1)额外内存空间复杂度下，原地删除数组中所有指定值元素的方法。通过一个指针跟踪不重复数据的位置，遍历数组并将不等于目标值的元素前移，最终返回新长度。这种方法不使用额外的储存空间，修改输入数组即可完成任务。

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Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Tip:题目要求不得使用额外的储存空间，也就是空间复杂度为O(1),另外解释了为什么不需要返回数组

public class RemoveElementSolution {
    public int removeDuplicates(int[] nums, int val) {
        /**
         * 此题与26很相似，改的只是判断条件
         */

        int newLen = 0;//不重复数据的指针
        int oldLen = nums.length;

        for (int i = 0; i < oldLen; i++) {
            if (nums[i] != val) {
                nums[newLen++] = nums[i];
            }

        }

        return newLen;
    }
}