Leetcode 26. Remove Duplicates from Sorted Array (Easy)

Leetcode Top Interview 150 的其中一題

題目描述

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
• Return k.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}
If all assertions pass, then your solution will be accepted.

範例

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints

• 1 <= nums.length <= 3 * 10^4
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

解法

解法一

解題思路：

把出現第二次以上的數字更改為 101 ，並將 duplicate +1 。因為nums[i]最大為 100，101一定會在排序時被排到最後面 ，所以前 len(nums) - duplicate 個數字一定都只出現一次

func removeDuplicates(nums []int) int {
    prev := nums[0]
    duplicate := 0

    for i:=1; i<len(nums); i++ {
        if nums[i] == prev {
            nums[i] =101
            duplicate++
        } else {
            prev = nums[i]
        }
    }

    sort.Ints(nums)

    return len(nums) - duplicate
}