Leetcode—67.二进制求和【简单】

2023每日刷题（二十八）

Leetcode—67.二进制求和

实现代码

void reverse(char *a, int len) {
    for(int i = 0; i < len / 2; i++) {
        char tmp = a[i];
        a[i] = a[len - 1 - i];
        a[len - 1 - i] = tmp;
    }
}

char* addBinary(char* a, char* b) {
    int len1 = strlen(a);
    int len2 = strlen(b);
    reverse(a, len1);
    reverse(b, len2);
    
    int maxn = len1 > len2 ? len1: len2;
    char *ans = (char *)malloc(sizeof(char) * (maxn + 2));
    int carry = 0;
    int i = 0, j = 0;
    int k = 0;
    while(i < len1 && j < len2) {
        int tmp = a[i++] - '0' + b[j++] - '0' + carry;
        ans[k++] = tmp % 2 + '0';
        carry = tmp / 2;
    }
    while(i < len1) {
        int tmp = a[i++] - '0' + carry;
        ans[k++] = tmp % 2 + '0';
        carry = tmp / 2;
    }
    while(j < len2) {
        int tmp = b[j++] - '0' + carry;
        ans[k++] = tmp % 2 + '0';
        carry = tmp / 2;
    }
    if(carry) {
        ans[k++] = '1';
    }
    ans[k] = '\0';
    reverse(ans, k);
    return ans;
}

运行结果

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