README_EN.md

27. Remove Element

中文文档

Description

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

Solutions

Approach 1: One Pass

We use the variable $k$ to record the number of elements that are not equal to $val$.

Traverse the array $nums$, if the current element $x$ is not equal to $val$, then assign $x$ to $nums[k]$, and increment $k$ by $1$.

Finally, return $k$.

The time complexity is $O(n)$ and the space complexity is $O(1)$, where $n$ is the length of the array $nums$.

Python3

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        k = 0
        for x in nums:
            if x != val:
                nums[k] = x
                k += 1
        return k

Java

class Solution {
    public int removeElement(int[] nums, int val) {
        int k = 0;
        for (int x : nums) {
            if (x != val) {
                nums[k++] = x;
            }
        }
        return k;
    }
}

C++

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int k = 0;
        for (int x : nums) {
            if (x != val) {
                nums[k++] = x;
            }
        }
        return k;
    }
};

Go

func removeElement(nums []int, val int) int {
	k := 0
	for _, x := range nums {
		if x != val {
			nums[k] = x
			k++
		}
	}
	return k
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function (nums, val) {
    let k = 0;
    for (const x of nums) {
        if (x !== val) {
            nums[k++] = x;
        }
    }
    return k;
};

Rust

impl Solution {
    pub fn remove_element(nums: &mut Vec<i32>, val: i32) -> i32 {
        let mut k = 0;
        for i in 0..nums.len() {
            if nums[i] != val {
                nums[k] = nums[i];
                k += 1;
            }
        }
        k as i32
    }
}

PHP

class Solution {

    /**
     * @param Integer[] $nums
     * @param Integer $val
     * @return Integer
     */
    function removeElement(&$nums, $val) {
        for ($i = count($nums) - 1; $i >= 0; $i--) {
            if ($nums[$i] == $val) {
                array_splice($nums, $i, 1);
            }
        }
    }
}

...