Valid Parentheses：括号匹配

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.

思路：用个栈匹配即可，注意最后若栈不空则不是完全匹配的，同理，若中途栈已空但仍试图匹配括号时，也是不能完全匹配的。

class Solution {   public boolean isValid(String s) {Stack<Character> st = new Stack<Character>();for (int i = 0; i < s.length(); i++) {if (s.charAt(i) == '(' || s.charAt(i) == '{' || s.charAt(i) == '[') {st.push(s.charAt(i));}if (s.charAt(i) == ')') {try {char c = st.peek();if (c == '(') {st.pop();} else {return false;}} catch (Exception e) {return false;}}if (s.charAt(i) == ']') {try {char c = st.peek();if (c == '[') {st.pop();} else {return false;}} catch (Exception e) {return false;}}if (s.charAt(i) == '}') {try {char c = st.peek();if (c == '{') {st.pop();} else {return false;}} catch (Exception e) {return false;}}}if (st.isEmpty()){return true;} else {return false;}   }}

当然，代码不用这么长，因为只包含括号，所以可以这么写：

public boolean isValid(String s) {Stack<Character> stack = new Stack<Character>();for (char c : s.toCharArray()) {if (c == '(')stack.push(')');else if (c == '{')stack.push('}');else if (c == '[')stack.push(']');else if (stack.isEmpty() || stack.pop() != c)return false;}return stack.isEmpty();}

不够匹配或者不能匹配即返回FALSE，还是很省事的。