Question

我的LeetCode上的反向整数代码不被接受。

我检查了程序是否可以返回正确答案。

class Solution:
    def reverse(self, x: int) -> int:
        check_num = str(x)
        flag = 0
        if(check_num[0] == '-'):
            check_num = check_num[1:]
            flag = 1

        elif (check_num[len(check_num)-1] == '0'):
            check_num = check_num[:len(check_num)-1]

        #print(check_num)

        #reverse
        time = len(check_num)
        storage = [0] * time 
        for i in range(len(check_num)):
            num = len(check_num)-i-1
            storage[i] = check_num[num]
            #print(storage[i])

        if(flag == 1):
            storage.insert(0, '-')

        #turn to string
        oneLinerString=""
        for x in storage:
            oneLinerString += x

        ans = int(oneLinerString)

        return oneLinerString

def main():
    import sys
    import io
    def readlines():
        for line in io.TextIOWrapper(sys.stdin.buffer, encoding='utf-8'):
            yield line.strip('\n')

    lines = readlines()
    while True:
        try:
            line = next(lines)
            x = int(line);

            ret = Solution().reverse(x)

            out = str(ret);
            print(out)
        except StopIteration:
            break

if __name__ == '__main__':
    main()

对于一种输入方式，我的程序返回正确的输出。

Your input
123
Output
321
Expected
321

但是有错误，我的代码不被接受。有什么问题，我该如何解决当前的代码？

Finished in N/A
ValueError: invalid literal for int() with base 10: ''
Line 30 in reverse (Solution.py)
Line 47 in main (Solution.py)
Line 55 in <module> (Solution.py)

Answer 1

对于输入0，由于以下原因，您的代码会将输入转换为空字符串：

elif (check_num[len(check_num)-1] == '0'):
            check_num = check_num[:len(check_num)-1]

您应该删除此elif分支，并让最终的整数转换处理反转数字的前导零：

ans = int(oneLinerString)  # removes leading zeros in the reversed string

当反转的数字超出32位有符号整数表示的范围时，还需要注意有关返回0的条件。因此，可以添加最终检查：

if not -2**31 <= ans <= 2**31 - 1:
    return 0

对示例代码进行最小的更改，一个可行的解决方案是：

class Solution:
    def reverse(self, x: int) -> int:
        check_num = str(x)
        flag = 0
        if(check_num[0] == '-'):
            check_num = check_num[1:]
            flag = 1

        #print(check_num)

        #reverse
        time = len(check_num)
        storage = [0] * time 
        for i in range(len(check_num)):
            num = len(check_num)-i-1
            storage[i] = check_num[num]
            #print(storage[i])

        if(flag == 1):
            storage.insert(0, '-')

        #turn to string
        oneLinerString=""
        for x in storage:
            oneLinerString += x

        ans = int(oneLinerString)  # removes leading zeros in the reversed string

        if not -2**31 <= ans <= 2**31 - 1:
            return 0

        return ans

Answer 2

尝试使用python的优势：

def solve():
    n = input()
    l = list(map(int,str(n)))
    l = l[::-1]

    #removing leading zeros
    i = 0
    while l[i] == 0:
        i+=1
    l = l[i:]
    n = ('').join(str(x) for x in l)
    return int(n)

if __name__ == "__main__":
    print (solve())

Answer 3

我已将 int 转换为字符串，然后将字符串转换为列表。我可以轻松地反转列表，然后可以按照相同的过程将列表再次转换为 int 对象。

num=-1534236469
sign=1
if num<0:
    sign=-1
    num=num*-1 # make positive for reverse
list1=list(str(num))
list1.reverse()

numrev=int("".join(str(x) for x in list1))*sign
if numrev.bit_length()>31:
    numrev=0

与传统的模除法相比，该算法将花费更多的时间。

正确答案，但运行时错误反向整数-LeetCode

3 个答案: