【Leet Code】20. Valid Parentheses
【Leet Code】20. Valid Parentheses
韩旭051
发布于 2019-11-08 00:42:25
发布于 2019-11-08 00:42:25
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本文链接：https://blog.csdn.net/shiliang97/article/details/101993070
Given a string 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.
Note that an empty string is also considered valid.
Example 1:
Input: "()"
Output: trueExample 2:
Input: "()[]{}"
Output: trueExample 3:
Input: "(]"
Output: falseExample 4:
Input: "([)]"
Output: falseExample 5:
Input: "{[]}"
Output: true简单题。。但是还是看的柳婼的答案过的。。用的 栈hh
class Solution {
public:
bool isValid(string s) {
stack<int> t;
for(int i=0;i<s.length();i++)
{
if(s[i]=='('||s[i]=='['||s[i]=='{'){
t.push(s[i]);
}else if(s[i]==')'){
if(i==0||t.empty()||t.top()!='('){
return false;
}
t.pop();
}else if(s[i]=='}'){
if(i==0||t.empty()||t.top()!='{'){
return false;
}
t.pop();
}else if(s[i]==']'){
if(i==0||t.empty()||t.top()!='['){
return false;
}
t.pop();
}
}
return t.empty();
}
};if(i==0||t.empty()||t.top()!='(')学习这个判断就行了~
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原始发表：2019-10-03 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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