Leetcode 每日一题C 语言版 -- 26 remove duplicates from sorted array

Leetcode 每日一题C 语言版 – 26 remove duplicates from sorted array

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问题描述

https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/?envType=study-plan-v2&envId=top-interview-150

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.

Consider the number of unique elements in nums to be k​​​​​​​​​​​​​​. After removing duplicates, return the number of unique elements k.

The first k elements of nums should contain the unique numbers in sorted order. The remaining elements beyond index k - 1 can be ignored.

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.

 

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 * 104
-100 <= nums[i] <= 100
nums is sorted in non-decreasing order.

解法一

双指针思想经典应用，slow 不等于fast 时，nums[++slow] = nums[fast]

int removeDuplicates(int* nums, int numsSize) {
    int slow = 0, fast = 1;

    if (numsSize <= 1)
        return numsSize;

    while (fast < numsSize) {
        if (nums[slow] != nums[fast])
            nums[++slow] = nums[fast];
        fast++;
    }

    return ++slow;
}