LeetCode 27 remove element
题目：
Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.
Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:
Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.
Return k.
Custom Judge:
The judge will test your solution with the following code:
int[] nums = […]; // Input array
int val = …; // Value to remove
int[] expectedNums = […]; // The expected answer with correct length.
// It is sorted with no values equaling val.
int k = removeElement(nums, val); // Calls your implementation
assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
assert nums[i] == expectedNums[i];
}
If all assertions pass, then your solution will be accepted.
Example 1:
Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,,]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).
Example 2:
Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,,,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).
Constraints:
0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100
解题思路：
- 首先，建立一个pointer k。我们用k来记录有多少个数字是不等于val的。k也会作为nums的index。
- 遍历nums中的每一个数字。如果这个数字不等于val，我们就把它移到index=k的位置，这样到最后，nums里面的前k个数字也全部都是不等于val的数字。
- 当我们把一个数字放在index=k的位置后，需要更新k，即是把k加1。
- 遍历结束后，k代表nums里面有多少个数不等于val。nums里面的前k个数字也全部都是不等于val的数字。
class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
k = 0
for i in range(len(nums)):
if nums[i] != val:
nums[k] = nums[i]
k += 1
return k
原文地址：https://blog.csdn.net/weixin_57266891/article/details/142697816
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