python leetcode 27. Remove Element

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本文介绍了一种在不使用额外空间的情况下，从数组中移除所有指定数值的方法，并通过修改输入数组原地完成，同时返回新的数组长度。示例展示了如何在保持前部元素正确的同时，忽略后续元素的状态。

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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.

去除数组中的指定数字，返回长度。（最后数组中药留下正确的数字才行，不可只计算剩下数组的长度。）

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]);
}

class Solution:
def removeElement(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
for num in nums[:]:
if num == val:
nums.remove(num)
return len(nums)