题目描述
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.
GoLang 语法：方法
- 方法的定义
func (variable_name variable_data_type) function_name() [return_type]{
// function body
}（1）变量 variable_name 被称为 接受者(Receiver)
（2）表达式 variable_name .function_name 被称为 Selector
（3）variable_name .function_name 是 method value 类型，跟普通函数类型一样。而 variable_data_type.function_name 是 method expression 类型，它类似于 C++ 的成员函数指针。
2. nil 也可以作为接受者
AC 代码
type array []int
func (arr *array) removeElement(val int){
newLen := 0
for i := 0; i < len(*arr); i++ {
if (*arr)[i] != val {
(*arr)[newLen] = (*arr)[i]
newLen++
}
}
*arr = (*arr)[:newLen]
}
func removeElement(nums []int, val int) int {
var arr array = nums
arr.removeElement(val)
return len(arr)
}
