[Swift]LeetCode80. 刪除排序數組中的重複項 II | Remove Duplicates from Sorted Array II

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Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twiceand 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 = [1,1,1,2,2,3],

Your function should return length = , with the first five elements of  being  and 3 respectively.

It doesn't matter what you leave beyond the returned length.5nums1, 1, 2, 2

Example 2:

Given nums = [0,0,1,1,1,1,2,3,3],

Your function should return length = , with the first seven elements of  being modified to , 0, 1, 1, 2, 3 and 3 respectively.

It doesn't matter what values are set beyond the returned length.
7nums0

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.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

---

給定一個排序數組，你須要在原地刪除重複出現的元素，使得每一個元素最多出現兩次，返回移除後數組的新長度。

不要使用額外的數組空間，你必須在原地修改輸入數組並在使用 O(1) 額外空間的條件下完成。

示例 1:

給定 nums = [1,1,1,2,2,3],

函數應返回新長度 length = , 而且原數組的前五個元素被修改成  3 。

你不須要考慮數組中超出新長度後面的元素。51, 1, 2, 2,

示例 2:

給定 nums = [0,0,1,1,1,1,2,3,3],

函數應返回新長度 length = , 而且原數組的前五個元素被修改成 , 0, 1, 1, 2, 3, 3 。

你不須要考慮數組中超出新長度後面的元素。
70

說明:

爲何返回數值是整數，但輸出的答案是數組呢?

請注意，輸入數組是以「引用」方式傳遞的，這意味着在函數裏修改輸入數組對於調用者是可見的。

你能夠想象內部操做以下:

// nums 是以「引用」方式傳遞的。也就是說，不對實參作任何拷貝
int len = removeDuplicates(nums);

// 在函數裏修改輸入數組對於調用者是可見的。
// 根據你的函數返回的長度, 它會打印出數組中該長度範圍內的全部元素。
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

---

12ms

 1 class Solution {
 2     func removeDuplicates(_ nums: inout [Int]) -> Int {
 3         if nums.count == 0 { return 0 }
 4         var result = 0
 5         
 6         var j = 1
 7         var repeatCount = 0
 8         var pre = nums[0]
 9         for i in 1..<nums.count {
10             if nums[i] == pre {
11                 repeatCount += 1
12                 if repeatCount <= 1 {
13                     nums[j] = nums[i]
14                     j += 1
15                 }
16             } else {
17                 pre = nums[i]
18                 repeatCount = 0
19                 nums[j] = nums[i]
20                 j += 1
21             }
22         }
23         return j
24     }
25 }

---

16ms

 1 class Solution {
 2     func removeDuplicates(_ nums: inout [Int]) -> Int {
 3         guard !nums.isEmpty else {return 0}
 4         var ip = 0, currCount = 1
 5         for jp in 1..<nums.count {
 6             if nums[ip] == nums[jp] {
 7                 currCount += 1
 8             } else {
 9                 currCount = 1
10             }
11             if currCount <= 2 {
12                 ip += 1
13                 (nums[ip], nums[jp]) = (nums[jp], nums[ip])
14             }
15         }
16         return ip + 1
17     }
18 }

---

40ms

 1 class Solution {
 2     func removeDuplicates(_ nums: inout [Int]) -> Int {
 3         var i = 0
 4         for num in nums {
 5             if i < 2 || num > nums[i-2] {
 6                 nums[i] = num
 7                 i += 1
 8             }
 9         }
10         nums[i...] = []
11         return i
12     }
13 }

---

44ms

 1 class Solution {
 2     func removeDuplicates(_ nums: inout [Int]) -> Int {
 3         guard nums.count > 0 else {
 4             return 0
 5         }
 6         var position = 1 // [0..<position]: result array
 7         var pre = nums[0] // previous value to compare with
 8         var repeatCount = 0
 9         for i in 1 ..< nums.count {
10             if nums[i] == pre {
11                 repeatCount += 1
12                 if repeatCount <= 1 {
13                     // repeat count less than 2, put the value into the result position
14                     nums[position] = nums[i]
15                     position += 1
16                 }
17             } else {
18                 repeatCount = 0
19                 // not equal to previous value, put it into the result position
20                 pre = nums[i]
21                 nums[position] = nums[i]
22                 position += 1
23             }
24         }
25         nums[position...] = []
26         return position
27     }
28 }