题目描述
Given a sorted array nums, remove the duplicates in-place such that each element appear only once 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.
Example 1:
Given nums = [1,1,2],
Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,0,1,1,1,2,2,3,3,4],
Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
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.
解题思路
这个问题要解决并不难, 但是题目要求不能重新申请空间,并且要求时间复杂度为 O(1), 就需要特殊的处理方式 并且题目中提示了我们只需要返回数组长度,并且验证的时候是只验证数组长度的值, 所以我们可以直接遍历数组, 将后边不同值直接覆盖前边重复的值
时间复杂度: O(n)
示例代码
func removeDuplicates(_ nums: inout [Int]) -> Int {
if (nums.count < 2) {
return nums.count
}
var index = 0
for i in 1..<nums.count {
if nums[i] != nums[index] {
index += 1
nums[index] = nums[i]
}
}
return index+1
}
