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LeetCode/remove-duplicates-from-sorted-array
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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]); | |
} | |
My code: | |
class Solution: | |
def removeDuplicates(self, nums): | |
""" | |
:type nums: List[int] | |
:rtype: int | |
""" | |
count=0 | |
i=0 | |
while i<(len(nums)): | |
if i==0: | |
count+=1 | |
else: | |
if nums[i]==nums[i-1]: | |
nums.remove(nums[i-1]) | |
continue | |
count+=1 | |
i+=1 | |
return count | |
Solution: | |
https://leetcode.com/problems/remove-duplicates-from-sorted-array/ |