LeetCode //C - 67. Add Binary

67. Add Binary

Given two binary strings a and b , return their sum as a binary string.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

Constraints:
  • 1 < = a . l e n g t h , b . l e n g t h < = 1 0 4 1 <= a.length, b.length <= 10^4 1<=a.length,b.length<=104
  • a and b consist only of '0' or '1' characters.
  • Each string does not contain leading zeros except for the zero itself.

From: LeetCode

Link: 67. Add Binary


Solution:

Ideas:
  1. Determine the Lengths: The lengths of the input strings a and b are determined using strlen.

  2. Allocate Memory: Memory is dynamically allocated for the result string, which must be long enough to hold the sum of the two binary strings plus a possible carry bit at the beginning and a null terminator at the end.

  3. Initialize Result: The allocated result string is initialized with '0's to prepare for the addition operation. This also handles edge cases where the input strings are "0".

  4. Perform Binary Addition:

  • Starting from the rightmost bit (least significant bit) of both strings, each bit is added along with any carry from the previous addition.
  • The sum of the two bits and the carry is computed. The resulting bit (either '0' or '1') is stored in the corresponding position in the result string, and the carry (either 0 or 1) is updated for the next iteration.
  • The loop continues, moving left through the strings until all bits have been processed.
  1. Handle Final Carry: After the loop, if there is a remaining carry, it is placed at the beginning of the result string.

  2. Adjust Result:

  • If there is no carry and the result starts with '0', this leading '0' is unnecessary (except in the case where the result is actually "0"), so it is removed. This is done by creating a copy of the string starting from result[1] and freeing the original result.
  • If there is a carry or the result does not start with '0', the result is returned as is.
Code:
c 复制代码
char* addBinary(char* a, char* b) {
    int lengthA = strlen(a);
    int lengthB = strlen(b);
    int maxLen = lengthA > lengthB ? lengthA : lengthB;
    char* result = (char*)malloc(maxLen + 2); // +1 for possible carry, +1 for '\0'
    
    if (result == NULL) {
        return NULL; // Handle allocation failure if needed
    }
    
    // Initialize the result with '0's to handle the case where a and b are "0"
    for (int i = 0; i < maxLen + 2; i++) {
        result[i] = '0';
    }
    
    int carry = 0; // Initialize the carry
    result[maxLen + 1] = '\0'; // Null-terminate the result
    int i = lengthA - 1;
    int j = lengthB - 1;
    int k = maxLen; // Start from the end of the result

    // Add each bit from right to left
    while (i >= 0 || j >= 0 || carry) {
        int sum = carry;
        if (i >= 0) {
            sum += a[i] - '0'; // Convert from char to int and add to sum
            i--;
        }
        if (j >= 0) {
            sum += b[j] - '0'; // Convert from char to int and add to sum
            j--;
        }
        carry = sum / 2; // Update carry
        result[k] = (sum % 2) + '0'; // Determine the bit to store as char
        k--;
    }

    // Check if there is a carry that used the extra space at the beginning
    if (carry) {
        result[0] = '1';
        return result;
    } else {
        // If the first character is '0' and there was no carry, we need to skip it
        if (result[0] == '0') {
            char* shifted_result = strdup(result + 1);
            free(result);
            return shifted_result;
        } else {
            // If the first character is not '0', we just return the result as is
            return result;
        }
    }
}
相关推荐
蒋星熠13 分钟前
Flutter跨平台工程实践与原理透视:从渲染引擎到高质产物
开发语言·python·算法·flutter·设计模式·性能优化·硬件工程
小莞尔19 分钟前
【51单片机】【protues仿真】基于51单片机宠物投食系统
c语言·stm32·单片机·嵌入式硬件·51单片机
小欣加油39 分钟前
leetcode 面试题01.02判定是否互为字符重排
数据结构·c++·算法·leetcode·职场和发展
3Cloudream43 分钟前
LeetCode 003. 无重复字符的最长子串 - 滑动窗口与哈希表详解
算法·leetcode·字符串·双指针·滑动窗口·哈希表·中等
王璐WL1 小时前
【c++】c++第一课:命名空间
数据结构·c++·算法
空白到白2 小时前
机器学习-聚类
人工智能·算法·机器学习·聚类
小马学嵌入式~2 小时前
嵌入式 SQLite 数据库开发笔记
linux·c语言·数据库·笔记·sql·学习·sqlite
索迪迈科技2 小时前
java后端工程师进修ing(研一版 || day40)
java·开发语言·学习·算法
zzzsde2 小时前
【数据结构】队列
数据结构·算法
芒克芒克2 小时前
LeetCode 面试经典 150 题:删除有序数组中的重复项(双指针思想解法详解)
算法