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Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same.
Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.
Return k after placing the final result in the first k slots of nums.
Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.
Custom Judge:
The judge will test your solution with the following code:
代码语言：javascript复制Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).思路
遍历数组，只有后面的数比前面的数大，就说明前面的元素出现了一次。然后做一个累加值
因为是排序好了的数组，所以 num[n 1] >= num[n]
代码语言：javascript复制func removeDuplicates(nums []int) int {
if len(nums) == 0 {
return 0
}
i, j := 1, 0
for ; i < len(nums); i {
if nums[i] > nums[j] {
j
nums[j] = nums[i]
}
}
j
return j
}快慢指针做法
代码语言：javascript复制func removeDuplicates(nums []int) int {
slow := 0
for fast :=0; fast < len(nums);fast {
if nums[fast] != nums[slow] {
nums[slow 1] = nums[fast]
slow
}
}
return slow 1
}func removeDuplicates2(nums []int) int {
for i := 0; i 1 < len(nums); {
if nums[i] == nums[i 1] {
nums = append(nums[:i], nums[i 1:]...)
} else {
i
}
}
return len(nums)
}
