【每日一题1020】失踪的数字

:woman_mage: 给定一个连续的数字列表,元素从数字1开始。但是发现其中有且仅有一个数字失踪了。请编写一个函数,找出这个失踪的数字。

示例:
输入:[1, 3, 4],输出:2
输入:[1, 2, 3],输出:4

题目难度:中等
题目来源:codewars: Number Zoo Patrol

def missing_number(nums: list) -> int:
    pass

assert missing_number([2, 3, 4]) == 1
assert missing_number([1, 2, 4]) == 3
assert missing_number([1, 2, 3]) == 4
def missing_number(nums: list) -> int:
    i = 1
    for n in nums:
        if n != i:
            return i
        i += 1
    return i


assert missing_number([2, 3, 4]) == 1
assert missing_number([1, 2, 4]) == 3
assert missing_number([1, 2, 3]) == 4
1 Like

image

def missing_number(nums: list) -> int:
    return [i for i in range(1,len(nums) +2) if i not in nums][0]

assert missing_number([2, 3, 4]) == 1
assert missing_number([1, 2, 4]) == 3
assert missing_number([1, 2, 3]) == 4
from functools import reduce

def missing_number(nums: list) -> int:
    nums.extend(list(range(1, len(nums)+2)))
    data = reduce(lambda x,y:x ^ y, nums)
    return data

assert missing_number([2, 3, 4]) == 1
assert missing_number([1, 2, 4]) == 3
assert missing_number([1, 2, 3]) == 4

思路:利用位运算的异或运算解决。

  1. 首先,创造一个辅助列表,包含的是所有完整元素。
  2. 然后,将这个列表的元素统统添加到缺少独苗到原列表。
  3. 最后,利用归约函数reduce,将所有的元素互相进行异或^运算,因为它不是成对出现的,最终就会剩下缺失的数字了。

异或运算:当且仅当1和0之间运算时返回1,其他情况(0和0,1和1)都返回0。
规律:(1)任何数字异或自己,结果一定是0;(2)任何数字异或0,结果一定是数字自身。

@Mrhjs 同学的题解很棒!为了方便社区交流,后续题解建议使用下面这种markdown格式哦:

```python
你的题解代码
```

参考题解:

def missing_number(nums: list) -> int:
    return sum(range(1,len(nums)+2))-sum(nums)

assert missing_number([2, 3, 4]) == 1
assert missing_number([1, 2, 4]) == 3
assert missing_number([1, 2, 3]) == 4

思路:差值法。通过构建一个完整元素的辅助列表,然后通过比较两个列表求和结果,就能得出存在差异的那个数字。

3 Likes
def missing_number(nums: list) -> int:
    if nums[0] != 1:
        return 1
    else:
        for i in nums:
            if i + 1 not in nums:
                return i + 1


assert missing_number([2, 3, 4]) == 1
assert missing_number([1, 2, 4]) == 3
assert missing_number([1, 2, 3]) == 4
    public int missingNumber(ArrayList arrayList){
        for (int i = 1; i <= arrayList.size()+1; i++) {
            if (arrayList.indexOf(i) == -1){
                return i;
            }
        }
        return -1;
    }
def missing_number(nums: list) -> int:
    l = [i+1 for i in range(len(nums) + 1)]
    l2 =[i for i in l if i not in nums]
    return l2[0]


assert missing_number([2, 3, 4]) == 1
assert missing_number([1, 2, 4]) == 3
assert missing_number([1, 2, 3]) == 4
def missing_number(nums: list) -> int:
    index = 0
    while index<len(nums)-1:
        if nums[index]+1!=nums[index+1]:
            return nums[index]+1
        index+=1
    if nums[0]-1>0:
        return nums[0]-1
    else:
        return nums[-1]+1
def missing_number(nums: list) -> int:
    for i in range(1,len(nums)+1):
        if i not in nums: return i
    return len(nums)+1
def solution(nums: list):
    for i in range(0, len(nums)):
        if i == 0:
            if nums[0] > 1:
                return nums[0] - 1
            continue
        elif nums[i] == nums[i-1] + 1:
            continue
        else:
            return nums[i] - 1
    return nums[-1] + 1


assert solution([2, 3, 4]) == 1
assert solution([1, 2, 4]) == 3
assert solution([1, 2, 3]) == 4

image

def missing_number(nums: list) -> int:
   if nums[0] != 1:
      return 1
   else:
      for i in range(len(nums)-1):
         if  nums[i] + 1 != nums[i+1]:
            return nums[i] + 1
      else:
         return nums[-1] + 1
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