leetcode: explore-array-21 从排序数组中删除重复项

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 = 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:

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

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.

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]);
}

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

显然去重后，元素个数为 5

nums 需要依次进行去重，且只能在 nums 上进行修改
而最终 nums 的前 4 位必须是：

[0,1,2,3,4, ...]，先后顺序也需要保持。

4后面之所以是省略号，是因为题目并不在乎后面的数字是什么。
It doesn't matter what values are set beyond the returned length.

// 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]);
}

class Solution:
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0

        i = 0
        j = 0
        while i < len(nums):
            f = nums[i]
            while i < len(nums) and nums[i] == f:
                i += 1
            nums[j] = f
            j += 1
        return j