思路:
1、将字符串的每个字符进行遍历
2、如果发现是左括号,那么将该字符压入到栈中
3、如果是右括号,先去存储好的栈顶找到相应的值
4、若栈为空返回false,若匹配,pop该左括号,若不匹配也返回false
5、最后看存储栈中的做括号是否都匹配上了,也就是栈最后为空,返回true,否则返回false
public class T3 { public static void main(String[] args) { System.out.println(isMatch("asdasdadasdasdsa")); System.out.println(isMatch("{ada(ada)asdas}")); System.out.println(isMatch("{asdasd(asdasdasd}asda)")); } static boolean isMatch(String s) { if (null == s) { return false; } //定义左右括号的对应关系 Map<Character, Character> bracket = new HashMap<>(); bracket.put('(', ')'); bracket.put('{', '}'); bracket.put('[', ']'); Stack<Character> stack = new Stack<>(); char[] chars = s.toCharArray(); for (Character temp : chars) { if (bracket.containsKey(temp)) { stack.push(temp); } else if (bracket.containsValue(temp)) { if (stack.isEmpty()) { return false; } if (bracket.get(stack.peek()).equals(temp)) { stack.pop(); } else { return false; } } } return stack.isEmpty(); } }
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