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LeetCode: Valid Parentheses 解题报告
Valid Parentheses
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.
SOLUTION1:
使用stack来解决的简单题目。所有的字符依次入栈
1. 遇到成对的括号弹栈，弹栈不成对返回false.
2. 栈为空只能压入左括号
3. 扫描完成时，栈应该为空，否则返回FALSE.
1 public class Solution { 2 public boolean isValid(String s) { 3 if (s == null || s.length() == 0) { 4 return true; 5 } 6 7 Stack<Character> stack = new Stack<Character>(); 8 int len = s.length(); 9 for (int i = 0; i < len; i++) {10 char c = s.charAt(i);11 if (stack.isEmpty()) {12 // 栈为空的时候，不能放右括号13 if (c == ‘]‘ || c == ‘)‘ || c == ‘}‘) {14 return false;15 }16 stack.push(c);17 continue;18 }19 20 // 栈不为空时，必须 要对应弹出相对应的括号21 if (c == ‘)‘ && stack.peek() == ‘(‘22 || c == ‘]‘ && stack.peek() == ‘[‘23 || c == ‘}‘ && stack.peek() == ‘{‘24 ) {25 stack.pop();26 // 或者继续压入左括号 27 } else if (c == ‘(‘ || c == ‘[‘ || c == ‘{‘) {28 stack.push(c);29 // 否则错误退出 30 } else {31 return false;32 }33 }34 35 return stack.isEmpty();36 }37 }
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LeetCode: Valid Parentheses 解题报告
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