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Copy path27.remove-element.py
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81 lines (78 loc) · 2.01 KB
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#
# @lc app=leetcode id=27 lang=python3
#
# [27] Remove Element
#
# https://leetcode.com/problems/remove-element/description/
#
# algorithms
# Easy (45.53%)
# Likes: 989
# Dislikes: 2137
# Total Accepted: 482.6K
# Total Submissions: 1.1M
# Testcase Example: '[3,2,2,3]\n3'
#
# Given an array nums and a value val, remove all instances of that value
# in-place 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.
#
# The order of elements can be changed. It doesn't matter what you leave beyond
# the new length.
#
# Example 1:
#
#
# Given nums = [3,2,2,3], val = 3,
#
# Your function should return length = 2, with the first two elements of nums
# being 2.
#
# It doesn't matter what you leave beyond the returned length.
#
#
# Example 2:
#
#
# Given nums = [0,1,2,2,3,0,4,2], val = 2,
#
# Your function should return length = 5, with the first five elements of nums
# containing 0, 1, 3, 0, and 4.
#
# Note that the order of those five elements can be arbitrary.
#
# 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 = removeElement(nums, val);
#
# // 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]);
# }
#
# @lc code=start
class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
i = 0
while i < len(nums):
if nums[i] == val:
nums.pop(i)
else:
i = i+1
return
# @lc code=end