20. Valid Parentheses（stack）

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.

public class Solution {

if (s == null)
			return false;
		Stack<Character> stack = new Stack<Character>();
		char[] ch = s.toCharArray();
		for (int i = 0; i < ch.length; i++) {
			if ((ch[i] == ‘(‘) || (ch[i] == ‘[‘) || (ch[i] == ‘{‘))
				stack.push(ch[i]);
			else {
				if (stack.isEmpty())
					return false;
				if (ch[i] - stack.pop() > 2)
					return false;
			}
		}
		return stack.isEmpty();

    public boolean isValid(String s) {
        if (s == null)                       //改进1：可以不用String.charat(i)
			return false;          //改进2： 先判断长度                                              //改进三： 不需要这么多if else ，符号的判断可以用ascii码
		Stack<Character> stack = new Stack<Character>();
		char[] ch = s.toCharArray();
		for (int i = 0; i < ch.length; i++) {
			if ((ch[i] == ‘(‘) || (ch[i] == ‘[‘) || (ch[i] == ‘{‘))
				stack.push(ch[i]);
			else {
				if (ch[i] == ‘)‘) {
					if (stack.isEmpty())
						return false;
					else if (stack.pop() != ‘(‘)
						return false;
				}
				if (ch[i] == ‘]‘) {
					if (stack.isEmpty())
						return false;
					if (stack.pop() != ‘[‘)
						return false;
				}
				if (ch[i] == ‘}‘) {
					if (stack.isEmpty())
						return false;
					if (stack.pop() != ‘{‘)
						return false;
				}
			}
		}
		if (!stack.isEmpty())
			return false;
		else
			return true;
    }
}