[LeetCode]Find All Duplicates in an Array

题目描述:

LeetCode 442. Find All Duplicates in an Array

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?

Example:

Input

[4,3,2,7,8,2,3,1]

Output

[2,3]

题目大意:

给定一个整数数组,1 <= a[i] <= n (n = 数组长度),某些元素出现两次,某些出现一次。寻找数组中所有出现两次的元素。你可以不使用额外空间并且在O(n)运行时间内完成题目吗?

解题思路:

解法I 正负号标记法(一趟遍历)

参考LeetCode Discuss:https://discuss.leetcode.com/topic/64735/java-simple-solution

遍历nums,记当前数字为n(取绝对值),将数字n视为下标(因为a[i]∈[1, n])

当n首次出现时,nums[n - 1]乘以-1

当n再次出现时,则nums[n - 1]一定<0,将n加入答案

Python代码:

class Solution(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        ans = []
        for n in nums:
            if nums[abs(n) - 1] < 0:
                ans.append(abs(n))
            else:
                nums[abs(n) - 1] *= -1
        return ans

解法II 位置交换法

遍历nums,记当前下标为i

当nums[i] > 0 并且 nums[i] != i + 1时,执行循环:

令n = nums[i]

如果n == nums[n - 1],则将n加入答案,并将nums[i]置为0

否则,交换nums[i], nums[n - 1]

Python代码:

class Solution(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        ans = []
        for i in range(len(nums)):
            while nums[i] and nums[i] != i + 1:
                n = nums[i]
                if nums[i] == nums[n - 1]:
                    ans.append(n)
                    nums[i] = 0
                else:
                    nums[i], nums[n - 1] = nums[n - 1], nums[i]
        return ans

 

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