Leetcode 27. Remove Element
11 Jul 2024leetcode easy array leetcode-top-interview Leetcode Top Interview 150 的其中一題
題目描述
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 nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums. Return k.
Custom Judge:
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.
範例
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 <= 100
- 0 <= nums[i] <= 50
- 0 <= val <= 100
解法
方法：檢查每個數字現在是否等於 val，是的話改成 100001，這樣排序後就會在最後面，不在列入檢查的陣列前面
func removeElement(nums []int, val int) int {
count := 0
for i:=0; i<len(nums); i++ {
if nums[i] == val {
nums[i] = 101
count++
}
}
sort.Ints(nums)
return len(nums)-count
}
Time complexity : O(nlogn)