本篇博文總結了《劍指Offer》面試題3：陣列中重複數字的幾種解法，以及LeetCode中與之相似的題目，旨在舉一反三能夠很好地解決這一類問題。

面試題3： 陣列中重複的數字

在一個長度為n的數組裡的所有數字都在0到n-1的範圍內。陣列中某些數字是重複的，但不知道有幾個數字是重複的，也不知道每個數字重複幾次。請找出陣列中任意一個重複的數字。例如，如果輸入長度為7的陣列{2,3,1,0,2,5,3}，那麼對應的輸出是第一個重複的數字2。

解法一： 建立一個長度為n的陣列dup，記錄每個數字出現的次數。時間複雜度為O(n)，需要O(n)的額外空間。

def duplicate(self, numbers, duplication)
 
:
        # write code here
        n = len(numbers)
        dup = [0] * n
        for i in numbers:
            dup[i] += 1
        for x in range(n):
            if dup[x] > 1:
                duplication[0] = x
                return True
        return False

解法二：

def duplicate(self, numbers, duplication)
 
:
        # write code here
        i = 0
        while i < len(numbers):
            if numbers[i] != i:
                if numbers[numbers[i]] == numbers[i]:
                    duplication[0] = numbers[i]
                    return True
                else:
                    tmp = numbers[numbers[i]]
                    numbers[numbers[i]] = numbers[i]
                    numbers[i] = tmp
            i += 1
 

        return False

LeetCode 287 （同面試題3中的題目2）： 不修改陣列找出重複數字

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Note:

• You must not modify the array (assume the array is read only).
• You must use only constant, O(1) extra space.
• Your runtime complexity should be less than O(n2)
• There is only one duplicate number in the array, but it could be repeated more than once

def findDuplicate(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        lo = 1
        hi = len(nums)-1
        while lo < hi:
            count = 0
            mid = lo + (hi - lo)//2
            for i in nums:
                if i >= lo and i <= mid:
                    count += 1
            if count > mid - lo + 1:
                hi = mid
            else:
                lo = mid + 1
        return lo

LeetCode 442： 找出陣列中所有的重複元素

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements that appear twice in this array.
Could you do it without extra space and in O(n) runtime?

解法一： 與面試題3中的解法二類似，但res中可能儲存了多次重複的元素，因此返回時利用了集合的屬性。

def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        res = []
        i = 0
        while i < len(nums):
            if nums[i] == i + 1:
                i += 1
            else:
                if nums[i] == nums[nums[i]-1]:
                    res.append(nums[i])
                    i += 1
                else:
                    tmp = nums[nums[i]-1]
                    nums[nums[i]-1] = nums[i]
                    nums[i] = tmp
        return list(set(res))

解法二： 當掃描至陣列中的元素i，若i為第一次出現，則將下標(i-1)處的元素變為負，即nums[i-1] = -nums[i-1]，所以當掃描至i，而下標(i-1)處的元素為負時，i即為重複元素。

def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        res = []
        for i in nums:
            i = abs(i)
            if nums[i-1] < 0:
                res.append(i)
            else:
                nums[i-1] = -nums[i-1]
        return res

LeetCode 448： 找出陣列中沒有出現的元素

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

解答： 同上題中的解法二類似，最後統計大於0的元素的下標。

def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        res = []
        for i in nums:
            index = abs(i)-1
            if nums[index] > 0:
                nums[index] = -nums[index]
        for i in range(len(nums)):
            if nums[i] > 0:
                res.append(i+1)
        return res