LeetCode-26.Remove Duplicates from Sorted Array | 删除排序数组中的重复项

// 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]);
}

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: 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:

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 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.

Constraints:
0 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
nums is sorted in ascending order.

//Go
func removeDuplicates(nums []int) int {
	for i:=len(nums)-1;i>0;i-- {
        if nums[i] == nums[i-1] {
            nums = append(nums[:i],nums[i+1:]...)
        }
    }

    return len(nums)
}

//Go
func removeDuplicates(nums []int) int {
    if len(nums) == 0 {  //考虑特殊情况
        return 0
    }
    i := 0; //慢指针
    for j := 1;j < len(nums);j++ {  //j是快指针
        if nums[j] != nums[i] {
            i++
            nums[i] = nums[j]
        }
    }
    return i + 1
}

leetcode-cn：
执行用时：8 ms, 在所有 Go 提交中击败了86.25%的用户
内存消耗：4.6 MB, 在所有 Go 提交中击败了65.76%的用户

leetcode：
Runtime: 4 ms, faster than 99.38% of Go online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 4.6 MB, less than 100.00% of Go online submissions for Remove Duplicates from Sorted Array.