Bracket matching
Given a string that only includes'(',')','{','}','[',']', determine whether the string is valid
[valid string must be satisfied]
1. The left parenthesis must be used The same type of closing parenthesis is closed.
2. The left parenthesis must be closed in the correct order.
3. Note that an empty string can be considered a valid string.
[Question Analysis]:
We know that any kind of parenthesis appears in pairs. Therefore, if the left parenthesis appears again, there must be a matching right parenthesis, but if only this condition is met, it cannot be performed normally. Matching, because it is not enough to make the number of occurrences the same, and it is also necessary to let them combine in the correct order. Therefore, we use the stack structure, use its first-in-last-out feature to push the left parenthesis in, compare the price of the right parenthesis with the top element of the stack, and pop it out of the stack. Finally, when the stack is empty, the traversal ends at the same time, indicating a match. The specific code is as follows:
first simply construct a stack
typedef char STDataType;
typedef struct stack
{
STDataType* _data;
int _size;
int _capacity;//栈的容量
}stack;
void initStack(stack* st)//初始化栈
{
if (st == NULL)
return;
st->_data = NULL;
st->_capacity = st->_size = 0;
}
void checkCapacity(stack* st)//容量检查
{
if (st->_capacity == st->_size)
{
int newCap = st->_capacity == 0 ? 1 : 2 * st->_capacity;
st->_data = (STDataType*)realloc(st->_data, sizeof(STDataType)*newCap);
st->_capacity = newCap;
}
}
void stackpush(stack* st, STDataType val)//压栈
{
if (st == NULL)
return;
checkCapacity(st);
st->_data[st->_size++] = val;
}
void stackpop(stack* st)//出栈
{
if (st == NULL || st->_size == 0)
return;
--st->_size;
}
STDataType stackTop(stack* st)//取栈顶元素
{
if (st == NULL || st->_size == 0)
return 0;
return st->_data[(st->_size) - 1];
}
int stackEmpty(stack* st)//判空
{
if (st == NULL || st->_size == 0)
return 1;
else
return 0;
}
On this basis, judge the subroutine:
//判断函数
bool isRight(char* s)
{
//括号映射
char map[3][2] = {
{
'(', ')' }, {
'{', '}' }, {
'[', ']' } };
struct stack st;
initStack(&st);
while (*s)
{
int flag = 0;
for (int i = 0; i < 3; ++i)//判断是否为左括号
{
if (*s == map[i][0])
{
stackpush(&st, *s);
//判断下一个字符
++s;
flag = 1;
break;
}
}
if (flag == 0)
{
//取出栈顶元素,先判断栈是否为空
if (stackEmpty(&st))
return false;
char topchar = stackTop(&st);
stackpop(&st);
for (int i = 0; i < 3; ++i)//判断与哪一个匹配
{
if (map[i][1] == *s)
{
if (map[i][0] == topchar)
{
//判断下一个字符
++s;
break;
}
else
return false;//没有正确匹配
}
}
}
}
//判断栈是否为空
return stackEmpty(&st);
}