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

题目

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

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 a 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]);
}

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.

题解

这道题就是希望让我返回一个没有重复数字的数组的大小，且不要使用额外的数组空间，必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。

解法一：暴力

用for循环从数组nums末尾的最后一个数字开始遍历，让数组nums末尾最后一个数字与末尾最后倒数第二个数字做比较，如果后一个等于前一个，则从数组中删除后一个元素，直到遍历完成，返回新数组的长度。

//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)
}

解法二：双指针法


//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
}

执行结果:

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.