原题链接
匹配括号
思路：
用栈，遍历过程中，匹配的成对出栈；结束后，栈空则对，栈非空则错。
Runtime: 4 ms, faster than 99.94% of Java
class Solution {public boolean isValid(String s) {Stack<Character> sta = new Stack<Character>();for (int i = 0; i < s.length(); i++) {char temp = s.charAt(i);if (temp == '[' || temp == '(' || temp == '{') {sta.push(temp);} else {if (sta.isEmpty())return false;char top = ' ';if (temp == ']')top = '[';if (temp == ')')top = '(';if (temp == '}')top = '{';if (top == sta.peek())sta.pop();elsereturn false;}}return sta.isEmpty() ? true : false;}}