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20. Valid Parentheses(stack)

Posted on 2016-01-27 13:16  cKK  阅读(191)  评论(0编辑  收藏  举报

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; } }