Valid Parentheses（48.24%）

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.

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效率低的（1.35%）

public class Solution {
public boolean isValid(String s) {
if (s.length() % 2 != 0) {
return false;
}

if (s.equals("")) {
return true;
} else {
if (s.contains("()")||s.contains("[]")||s.contains("{}")) {
String a = s.replaceAll("\\(\\)", "");
String b = a.replaceAll("\\[\\]", "");
String c = b.replaceAll("\\{\\}", "");
return isValid(c);
} else {
return false;
}
}
}
}

上面的做法虽然容易读懂，但是效率非常低，搜了一下，推荐的做法是：

利用一个栈来保存前括号，然后有后括号来时弹出栈顶来判断。

class Solution {
public:
bool isValid(string s) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
stack<char> st;

for(int i = 0; i < s.size(); i++)
if (s[i] == ')' || s[i] == ']' || s[i] == '}')
{
if (st.empty())
return false;
else
{
char c = st.top();
st.pop();
if ((c == '(' && s[i] != ')') || (c == '[' && s[i] != ']') || (c == '{' && s[i] != '}'))
return false;
}
}
else
st.push(s[i]);

return st.empty();
}
};

翻译成Java大概是这样（48.24%）：

import java.util.Stack;

public class Solution {
public boolean isValid(String s) {

char[] characters = s.toCharArray();
int len = characters.length;
if (len % 2 != 0) {
return false;
}

Stack<Character> lefts = new Stack<Character>();
for (int i = 0; i < len; i++) {
switch (characters[i]) {
case ')':
if (lefts.size() == 0 || lefts.pop() != '(')
return false;
break;
case ']':
if (lefts.size() == 0 || lefts.pop() != '[')
return false;
break;
case '}':
if (lefts.size() == 0 || lefts.pop() != '{')
return false;
break;

default:
lefts.add(characters[i]);
break;
}
}

return lefts.size() == 0;
}
}