C# 写 LeetCode easy #26 Remove Duplicates from Sorted Array

26、Remove Duplicates from Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appear only once 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.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

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

Example 2:

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

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

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

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

代码：

static void Main(string[] args)
{
    int[] nums = new int[] { ,,,,};
    int res = RemoveDuplicatesfromSortedArray(nums);
    Console.WriteLine(res);
    Console.ReadKey();
}

private static int RemoveDuplicatesfromSortedArray(int[] nums)
{
    if (nums == null || nums.Length == ) return ;
    int index = ;
    for (int i = ; i < nums.Length; i++)
    {
        if (nums[i] != nums[index])
        {
            index++;
            nums[index] = nums[i];
        }
    }
    return index+;
}

解析：

输入：整型数组

输出：不重复的个数

思想：

　　首先，对空数组或数组长度为0的输出结果为0。

　　其次，对数组中从第二个数开始和前面的数进行比较（这里前面的数是nums[index]，即去重后的最后一个数。开始时是第一个数），若不同，则将其存入到index索引下。

　　最后，由于第一个数之前没有计入，所以要加1。

时间复杂度：O(n)