Leetcode 20 Valid Parentheses
2016-08-30 16:52
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Given a string containing just the characters
determine if the input string is valid.
The brackets must close in the correct order,
all valid but
not.
括号匹配,书上讲栈应用的一个实例,新来的和栈顶元素比较就可以了。
好在这题并没有像之前的某些题目一样丧心病狂到出各种不合常理的输入数据。
class Solution {
public:
bool isValid(string s) {
stack<char> st;
for(int i=0;i<s.length();i++)
{
if(s[i]=='(' || s[i]=='[' || s[i]=='{')
st.push(s[i]);
else
{
if(st.empty()) return false;
if((s[i]==')' && st.top()=='(') || (s[i]==']' && st.top()=='[') || (s[i]=='}' && st.top()=='{'))
st.pop();
else
return false;
}
}
return st.empty();
}
};
'(',
')',
'{',
'}',
'['and
']',
determine if the input string is valid.
The brackets must close in the correct order,
"()"and
"()[]{}"are
all valid but
"(]"and
"([)]"are
not.
括号匹配,书上讲栈应用的一个实例,新来的和栈顶元素比较就可以了。
好在这题并没有像之前的某些题目一样丧心病狂到出各种不合常理的输入数据。
class Solution {
public:
bool isValid(string s) {
stack<char> st;
for(int i=0;i<s.length();i++)
{
if(s[i]=='(' || s[i]=='[' || s[i]=='{')
st.push(s[i]);
else
{
if(st.empty()) return false;
if((s[i]==')' && st.top()=='(') || (s[i]==']' && st.top()=='[') || (s[i]=='}' && st.top()=='{'))
st.pop();
else
return false;
}
}
return st.empty();
}
};
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