[LeetCode] 67. 二进制求和

题目链接 : https://leetcode-cn.com/problems/add-binary/

题目描述:

给定两个二进制字符串,返回他们的和(用二进制表示)。

输入为非空字符串且只包含数字 10

示例:

示例 1:

输入: a = "11", b = "1"
输出: "100"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

思路:

思路一: 库函数

二进制转十进制,十进制数相加再转二进制

class Solution:
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        return bin(int(a,2) + int(b,2))[2:]

思路二:模拟加法过程

代码:

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        res = ""
        carry = 0
        i = len(a) - 1
        j = len(b) - 1
        while i >=0 or j >= 0 or carry:
            tmp1 = int(a[i]) if i >= 0 else 0
            tmp2 = int(b[j]) if j >= 0 else 0
            carry, t = divmod(tmp1 + tmp2 + carry, 2)
            res = str(t) + res
            i -= 1
            j -= 1
        return res

java

class Solution {
    public String addBinary(String a, String b) {
        StringBuilder res = new StringBuilder();
        int i = a.length() - 1, j = b.length() - 1, carry = 0;
        while (i >= 0 || j >= 0 || carry != 0) {
            int tmp1 = (i >= 0 ? a.charAt(i--) - '0' : 0);
            int tmp2 = (j >= 0 ? b.charAt(j--) - '0' : 0);
            res.append((tmp1 + tmp2 + carry) % 2);
            carry = (tmp1 + tmp2 + carry) / 2;
        }
        return res.reverse().toString();      
    }
}

posted on 2019-05-27 15:18  威行天下  阅读(435)  评论(0编辑  收藏  举报

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