LeetCode 7:《Reverse Integer》

阿新 • • 發佈：2019-01-05

程式設計思路：注意處理overflow

class Solution:
    # @return an integer
    def reverse(self, x):
        result = ''
        abs_x = abs(x)
        # 整型轉字串，方便處理
        s = str(abs_x)
        for i in range(len(s)):
            result = result + s[len(s)-i-1]
        # 負數注意加上‘-’
        result_int = int(result) if x >= 0 else int(result)*(-1)
        # 處理Overflow
        if result_int > math.pow(2, 31) or result_int < math.pow(2, 31)*(-1):
            return 0
        return result_int

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寫法 art return 匯總 esc multipl 情況下 rip 簡單問題鏈接 LeetCode 7 題目解析給定一個32位有符號整數，求其反轉數字。解題思路如果是簡單反轉的話，那這道題就太簡單了。題目要求判斷溢出問題，32位int類型的範圍是-214748

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題目型別：字串題意：數字反轉，給出一個數字，將他的各個位反轉，例如123->321，-123->-321 - 注意負數符號不參與。 - 注意反轉後開頭的0要去掉，例如120->21 - 注意int反轉後可能成為double：1534236469/-1534

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LeetCode 7:《Reverse Integer》

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