# [LeetCode] 27. Remove Element 移除元素

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.

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]);
}```

Java:

```public class Solution {
public int removeElement(int[] nums, int val) {
int position = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != val) {
nums[position] = nums[i];
position++;
}
}
return position;
}
}　```

Python:

```class Solution(object):
def moveZero(self, nums, elem):
i = 0
for j in xrange(len(nums)):
if nums[j] != elem:
nums[i], nums[j] = nums[j], nums[i]
i += 1

return i　　```

Python:

```class Solution:
def removeElement(self, A, elem):
i, last = 0, len(A) - 1
while i <= last:
if A[i] == elem:
A[i], A[last] = A[last], A[i]
last -= 1
else:
i += 1
return last + 1
```

C++:

```class Solution {
public:
int removeElement(vector<int>& nums, int val) {
int res = 0;
for (int i = 0; i < nums.size(); ++i) {
if (nums[i] != val) nums[res++] = nums[i];
}
return res;
}
};
```

```void compressMe(char[] s, char c) {
int index = 0;
int i = 0;
for (; i < s.length; i++){
if (s[i] != c){
s[index++] = s[i];
}
}
for (; index < s.length; index++){
s[index] = ' ';
}
}
```

C++:

```private static void compressMe(char[] s, char c)
{
int insertIdx = 0;
for (int i = 0; i < s.Length; i++)
{
if (s[i] != c) s[insertIdx++] = s[i];
}

while (insertIdx < s.Length) s[insertIdx++] = ' ';
}
```

Python:  一个循环里

```class Solution(object):
def moveZero(self, nums, elem):
i = 0
for j in xrange(len(nums)):
if nums[j] != elem:
nums[i], nums[j] = nums[j], ''
i += 1
else:
nums[j] = ''
```

[LeetCode] 283. Move Zeroes 移动零

# All LeetCode Questions List 题目汇总

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### Leetcode练习第一题

ap.put(nums[i],i);}thrownewIllegalArgumentException("Notwosumsolution");}}作者：guanpengchn链接：https://leetcode-cn.com/problems/two-sum/solution/jie-suan-fa-1-liang-shu-zhi-he-by-guanpengchn/来源：力扣（LeetCod