Problem
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
numssuch that the firstkelements ofnumscontain the elements which are not equal toval. The remaining elements ofnumsare not important as well as the size ofnums. - 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 <= 1000 <= nums[i] <= 500 <= val <= 100
Approach
EN
Using splice, if current value(nums[i]) is equal to val(argument), remove the current value from the array and in that case, index has to be reduced as well.
- In the for loop, check each value with val and if it's equal to val, using
spliceto remove the value from the nums array. - Since the
nums[i+1]becamenums[i]due to splice,ihas to be reduced.(i--)
KR
반복문을 돌면서 nums 배열의 현재 원소와 아규먼트로 받은 val을 비교한다. 만약 같은 값이라면, splice를 이용해서 nums 배열에서 제거한다. 단, splice를 한 다음엔 기존의 nums[i+1] 가 nums[i]로 되기 때문에 반복문을 도는 인덱스 i를 감소시켜준다.
code
/**
* @param {number[]} nums
* @param {number} val
* @return {number}
*/
var removeElement = function(nums, val) {
for(let i=0; i<nums.length; i++) {
if (nums[i] === val) {
nums.splice(i, 1);
i--;
}
}
return nums.length;
};