#20 Valid Parentheses

最新推荐文章于 2020-02-26 11:26:29 发布

原创最新推荐文章于 2020-02-26 11:26:29 发布 · 520 阅读

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#LeetCode #C #String #Stack

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本文介绍了一个算法用于判断字符串中括号是否匹配并正确闭合，包括常见括号类型如(, ), {, }

题目链接：https://leetcode.com/problems/valid-parentheses/

Given a string containing just the characters '(', ')','{', '}', '[' and ']', determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

typedef struct StackRecord {    //不考虑溢栈实现简单的堆栈
	char ch[100];
	int top;
}*Stack;
bool isEmpty(Stack stack) {
	if (stack->top >= 0)
		return false;
	return true;
}
Stack init() {
	Stack stack = (Stack)malloc(sizeof(struct StackRecord));
	stack->top = -1;
	return stack;
}
void push(char ch, Stack stack) {
	stack->ch[++stack->top] = ch;
}
char top(Stack stack) {     //返回栈顶元素，空栈时返回'\0'
    if(!isEmpty(stack))
	    return stack->ch[stack->top];
	return 0;
}
void pop(Stack stack) {
	--stack->top;
}
bool isValid(char* s) {
	Stack stack = init();
	while (*s) {
		switch (*s) {
		case '(':       //左括号入栈
		case '[':
		case '{':
			push(*s, stack);
			break;
		case ')':      //右括号，查看栈顶是否为对应左括号。不是返回false；否则出栈
			if (top(stack) != '(')
				return false;
			pop(stack);
			break;
		case ']':
			if (top(stack) != '[')
				return false;
			pop(stack);
			break;
		case '}':
			if (top(stack) != '{')
				return false;
			pop(stack);
			break;
		default:
			return false;
		}
		++s;
	}
	if (!isEmpty(stack))        //栈内不空说明有多余左括号
		return false;
	return true;
}