# 645. Set Mismatch 错误的集合

## # 题目描述

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:

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

## # 解题方法

### # Hash方法

``````class Solution(object):
def findErrorNums(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
hashs = [0] * len(nums)
missing = -1
for i in range(len(nums)):
hashs[nums[i] - 1] += 1
return [hashs.index(2) + 1, hashs.index(0) + 1]
``````

### # 直接计算

268. Missing Numberopen in new window很像，直接计算出来也可以，速度稍快点。

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

## # 日期

2018 年 2 月 3 日 2018 年 11 月 21 日 —— 又是一个美好的开始