Add Binary
Total Accepted: 46815
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Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100"
.
我的解决方案:
class Solution { public: string addBinary(string a, string b) { string result = ""; int c = 0; int i = a.size() - 1; int j = b.size() - 1; while(i >= 0 || j >=0 ||c ==1) { c += i >= 0 ? a[i--] - '0':0; c += j >= 0 ? b[j--] - '0':0; result = char( c% 2 + '0') + result; c /= 2; } return result; } };
c语言解决方案:
char* addBinary(char* a, char* b) { int n, m; for (n=0; *a; a++, n++) ; for (m=0; *b; b++, m++) ; char *p = (char*)malloc(m>n ? m+2 : n+2), *last = p; int c = 0; while (n || m || c) { int s = c; if (n) { s += *(--a)-'0'; --n; } if (m) { s += *(--b)-'0'; --m; } *last++ = (s&1)+'0'; c = s>>1; } *last=0; char *start = p, t; while (start+1 < last) { // reverse string t = *start; *start++=*(--last); *last=t; } return p; }
数字电路版本代码:加法器的实现
https://leetcode.com/discuss/40846/c-4ms-solution-inspired-by-hardware-full-adder-circuit
python 的三个版本:
class Solution: # @param {string} a # @param {string} b # @return {string} def addBinary(self, a, b): i, m, n, result, carry = 1, len(a), len(b), [], 0 while i <= m or i <= n: temp = carry if i <= m: temp += int(a[-i]) if i <= n: temp += int(b[-i]) carry = temp / 2 result.append(str(temp % 2)) i += 1 if carry: result.append(str(carry)) return ''.join(result[::-1]) or a really short one if you want class Solution: # @param {string} a # @param {string} b # @return {string} def addBinary(self, a, b): return '{0:b}'.format(int(a, 2) + int(b, 2)) class Solution: # @param {string} a # @param {string} b # @return {string} def addBinary(self, a, b): return bin(int(a,2) + int(b,2))[2:]
时间: 2024-10-02 07:20:27