---
title: 'LeetCode 27. Remove Element'
disqus: hackmd
---

# LeetCode 27. Remove Element

## Description

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The relative order of the elements may be changed.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

## Example

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.

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.

## Constraints

0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100

## Answer

此題可將不是val的都重新給定一次，但遇到只有最後一個是val的情況，就會花較久時間。

int removeElement(int* nums, int numsSize, int val){
    int i , j = 0 , n = numsSize;
    if(n == 0){return 0;}
    if(n == 1 && nums[0] != val){return 1;}
    for(i = 0 ; i<n ; i++){
        if(nums[i] != val){nums[j] = nums[i];j++;}
    }
    return j;
}

我們可以改成，判斷有抓到第一個val確定要重新給定值得地方後，才會開始重新給定即可。最後的return條件是因位若全部都不是val的話，cnt仍為-1，則直接return numsSize即可。

//2022_03_29
int removeElement(int* nums, int numsSize, int val){
    int i = 0, cnt = -1;
    for(i = 0; i < numsSize; i++){
        if(cnt == -1){
            if(nums[i] == val){cnt = i;}
        }
        else{
            if(nums[i] != val){
                nums[cnt] = nums[i];
                cnt++;
            }
        }
    }
    return cnt != -1 ? cnt : numsSize;
}

## Link
https://leetcode.com/problems/remove-element/
###### tags: `Leetcode`