C# 写 LeetCode easy #27 Remove Element

27、 Remove Element

　　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.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.

代码：

static void Main(string[] args)
{
    int[] nums = { 0, 1, 3, 3, 4, 3, 5 };
    Console.WriteLine(RemoveElement(nums,3));
    Console.ReadKey();
}

private static int RemoveElement(int[] nums, int val)
{
    int index = 0;
    for(int i=0;i<nums.Length;i++)
    {
        if (nums[i] != val)
        {
            nums[index] = nums[i];
            index++;
        } 
    }
    return index;
}            

解析：

输入：数组和某个元素

输出：删除后的数组元素个数

代码思想：

　　用index记录在值不相同时的索引，循环遍历元素，若不相同将记录索引处的值用当前值覆盖，最后增加索引计数。

时间复杂度：O(n)