Given an array and a value, 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:
Given nums = [3,2,2,3], val = 3,Your function should return length = 2, with the first two elements of nums being 2.
关键
- Do not allocate extra space for another array.
- The order of elements can be changed.
- It doesn't matter what you leave beyond the new length.
方法一、直接去掉元素，前后两部分重新拼接。
func removeElement(nums []int, val int) int {
n := len(nums)
for i := 0; i < n; i++ {
if val == nums[i] {
nums = append(nums[:i], nums[i+1:]...)
i--
}
}
return len(nums)
}回过头看，slice类型的拼接，似乎额外申请空间了；虽然可以AC，但违规了，空间复杂度应该是O(n)。
方法二、利用原本数组空间重新“覆盖”一个新数组。
func removeElement(nums []int, val int) int {
j, n := 0, len(nums)
for i := 0; i < n; i++ {
if val != nums[i] {
nums[j] = nums[i]
j++
}
}
return j
}方法三、从尾部开始获取元素，也是“覆盖”一个新数组。
func removeElement(nums []int, val int) int {
n, i := len(nums), 0
for i < n {
if nums[i] == val {
nums[i] = nums[n-1]
n--
} else {
i++
}
}
return n
}