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

2六、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 = , with the first two elements of  being  and  respectively.

It doesn't matter what you leave beyond the returned length.2nums12

Example 2:

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

Your function should return length = , with the first five elements of  being modified to , , , , and  respectively.

It doesn't matter what values are set beyond the returned length.
5nums01234

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]);
}

代碼：

static void Main(string[] args)
{
    int[] nums = new int[] { 1,1,1,1,2};
    int res = RemoveDuplicatesfromSortedArray(nums);
    Console.WriteLine(res);
    Console.ReadKey();
}

private static int RemoveDuplicatesfromSortedArray(int[] nums)
{
    if (nums == null || nums.Length == 0) return 0;
    int index = 0;
    for (int i = 1; i < nums.Length; i++)
    {
        if (nums[i] != nums[index])
        {
            index++;
            nums[index] = nums[i];                   
        }
    }
    return index+1;           
}

 

解析：

輸入：整型數組

輸出：不重複的個數

思想：

　　首先，對空數組或數組長度爲0的輸出結果爲0。

　　其次，對數組中從第二個數開始和前面的數進行比較（這裏前面的數是nums[index]，即去重後的最後一個數。開始時是第一個數），若不一樣，則將其存入到index索引下。

　　最後，因爲第一個數以前沒有計入，因此要加1。

時間複雜度：O(n)