user: 这段汇编代码用C语言表示是什么样的:

```asm
<editDistance>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r13
  mov    %rsi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %eax,-0x50(%rbp)
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%r14d
  cmp    %eax,%ebx
  jge    L74
  mov    %r14d,-0x50(%rbp)
  mov    %ebx,%r14d
  mov    %r15,%rax
  mov    %r13,%r15
  mov    %rax,-0x48(%rbp)
L74:
  lea    0x1(%r14),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  lea    0xf(%rdx),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L123:
  cmp    %rcx,%rsp
  je     L121
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L123
L121:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L181
  orq    $0x0,-0x8(%rsp,%rax,1)
L181:
  mov    %rsp,%r13
  test   %r14d,%r14d
  js     L109
  lea    0x1(%r14),%ecx
  mov    $0x0,%eax
L37:
  mov    %eax,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L37
L109:
  lea    0xf(%rdx),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L176:
  cmp    %rdx,%rsp
  je     L83
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L176
L83:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L73
  orq    $0x0,-0x8(%rsp,%rax,1)
L73:
  mov    %rsp,%r12
  cmpl   $0x0,-0x50(%rbp)
  jle    L165
  movl   $0x1,-0x4c(%rbp)
  jmp    L1
L201:
  mov    -0x4(%r12,%rbx,4),%edx
  mov    0x0(%r13,%rbx,4),%esi
  mov    -0x4(%r13,%rbx,4),%edi
  call   <min@plt>
  add    $0x1,%eax
L14:
  mov    %eax,(%r12,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%r14d
  jl     L130
L127:
  mov    -0x48(%rbp),%rax
  movzbl -0x1(%rax,%rbx,1),%eax
  cmp    %al,(%r15)
  jne    L201
  mov    -0x4(%r13,%rbx,4),%eax
  jmp    L14
L130:
  mov    $0x0,%eax
L105:
  mov    (%r12,%rax,4),%edx
  mov    %edx,0x0(%r13,%rax,4)
  movl   $0x0,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r14d
  jge    L105
L33:
  addl   $0x1,-0x4c(%rbp)
  mov    -0x4c(%rbp),%eax
  add    $0x1,%r15
  cmp    %eax,-0x50(%rbp)
  jl     L165
L1:
  mov    -0x4c(%rbp),%eax
  mov    %eax,(%r12)
  test   %r14d,%r14d
  jle    L22
  mov    $0x1,%ebx
  jmp    L127
L22:
  jne    L33
  mov    (%r12),%eax
  mov    %eax,0x0(%r13)
  movl   $0x0,(%r12)
  jmp    L33
L165:
  movslq %r14d,%r14
  mov    0x0(%r13,%r14,4),%eax
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L146
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L146:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int editDistance(char a[], char b[]) {
  int aLength = strlen(a), bLength = strlen(b);
  int i, j;
  if (aLength < bLength) {
    char *tempPointer = a;
    a = b;
    b = tempPointer;
    int temp = aLength;
    aLength = bLength;
    bLength = temp;
  }
  int prev[bLength + 1];
  for (j = 0; j <= bLength; ++j) {
    prev[j] = j;
  }
  int curr[bLength + 1];
  for (i = 1; i <= aLength; ++i) {
    curr[0] = i;
    for (j = 1; j <= bLength; ++j) {
      if (a[i - 1] == b[j - 1]) {
        curr[j] = prev[j - 1];
      } else {
        curr[j] = 1 + min(prev[j - 1], prev[j], curr[j - 1]);
      }
    }
    for (j = 0; j <= bLength; ++j) {
      prev[j] = curr[j];
      curr[j] = 0;
    }
  }
  return prev[bLength];
}
```