Given a string s
containing just the characters '('
, ')'
, '{'
, '}'
, '['
and ']'
, determine if the input string is valid.
An input string is valid if:
- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
- Every close bracket has a corresponding open bracket of the same type.
Example 1:
**Input:** s = "()"
**Output:** true
Example 2:
**Input:** s = "()[]{}"
**Output:** true
Example 3:
**Input:** s = "(]"
**Output:** false
Constraints:
1 <= s.length <= 104
s
consists of parentheses only'()[]{}'
.
这道题让我们验证输入的字符串是否为括号字符串,包括大括号,中括号和小括号。这里需要用一个栈,开始遍历输入字符串,如果当前字符为左半边括号时,则将其压入栈中,如果遇到右半边括号时,若此时栈为空,则直接返回 false,如不为空,则取出栈顶元素,若为对应的左半边括号,则继续循环,反之返回 false,代码如下:
解法一:
class Solution {
public:
bool isValid(string s) {
stack<char> st;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(' || s[i] == '[' || s[i] == '{') {
st.push(s[i]);
} else {
if (st.empty()) return false;
if (s[i] == ')' && st.top() != '(') return false;
if (s[i] == ']' && st.top() != '[') return false;
if (s[i] == '}' && st.top() != '{') return false;
st.pop();
}
}
return st.empty();
}
};
再来看一种写的更加简洁的方法,这里用个小 trick,之前是当遇到左括号时,把左括号压入栈,这样在出栈检测的时候,得判断当前遇到的右括号是哪种括号,是否跟栈顶的左括号匹配。而假如在压入栈的时候,我们直接将左括号对应的右括号压入栈,则在出栈检测的时候,就直接可以比较是否和栈顶元素相等了,因为都是右括号,这样写起来能相对简单一点,但是整体的思路还是相同的,参见代码如下:
解法二:
class Solution {
public:
bool isValid(string s) {
stack<char> st;
for (char c : s) {
if (c == '(') st.push(')');
else if (c == '{') st.push('}');
else if (c == '[') st.push(']');
else if (st.empty() || st.top() != c) return false;
else st.pop();
}
return st.empty();
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/20
类似题目:
Different Ways to Add Parentheses
Check If Word Is Valid After Substitutions
Check if a Parentheses String Can Be Valid
Move Pieces to Obtain a String
参考资料:
https://leetcode.com/problems/valid-parentheses/
https://leetcode.com/problems/valid-parentheses/discuss/9178/Short-java-solution
LeetCode All in One 题目讲解汇总(持续更新中…)
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