leetcode 645. Set Mismatch——凡是要节约空间的题目 都在输入数据上下功夫 不要担心破坏原始的input

The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

Example 1:

Input: nums = [1,2,2,4]
Output: [2,3]

Note:

1. The given array size will in the range [2, 10000].
2. The given array's numbers won't have any order.

 

有三种解法，（1）直接sum计算找出缺失值，需要用到set找到重复数。（2）使用sort思路。数字按照预定的规则排列。（3）将set的使用放在nums里，nums里的数字为负数表示该位置存在对应数字。

代码分别如下：

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class Solution(object):     def findErrorNums(self, nums):         """         :type nums: List[int]         :rtype: List[int]         """         n = len(nums)         s = sum(set(nums))         return [sum(nums)-s, n*(n+1)/2-s]

 

 

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class Solution(object):     def findErrorNums(self, nums):         """         :type nums: List[int]         :rtype: List[int]         """         # sort         for i,n in enumerate(nums):             while nums[i] != nums[nums[i]-1]:                 t = nums[i]                                 nums[i] = nums[nums[i]-1]                 nums[t-1] = t         for i,n in enumerate(nums):             if n != i+1:                 return [n,i+1]

 

 

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class Solution(object):     def findErrorNums(self, nums):         """         :type nums: List[int]         :rtype: List[int]         """         for i in xrange(len(nums)):             pos = nums[i] if nums[i]>0 else -nums[i]             if nums[pos-1] > 0:                 nums[pos-1] = -nums[pos-1]             else:                 dup = pos         for i, n in enumerate(nums):             if n > 0:                 return [dup, i+1]