20. 有效的括号

 1     //方法一：使用栈，成对出现则弹栈，单个出现则入栈，如果是有效的对于括号，则栈最后为空
 2     public static boolean isValid10(String s) {
 3         if(s.isEmpty()) {
 4             return true;
 5         }
 6         //此判断可省略
 7         if(s.length() == 1) {
 8             return false;
 9         }
10         Map<Character, Integer> dict = new HashMap<>();
11         dict.put(‘(‘, 1);
12         dict.put(‘)‘, 2);
13         dict.put(‘[‘, 4);
14         dict.put(‘]‘, 5);
15         dict.put(‘{‘, 7);
16         dict.put(‘}‘, 8);
17         Stack<Character> stack = new Stack<>();
18         for(int i=0; i<s.length(); i++) {
19             if(stack.isEmpty()) {
20                 stack.push(s.charAt(i));
21                 continue;
22             }
23             if(dict.get(stack.peek()) == (dict.get(s.charAt(i))-1)) {
24                 stack.pop();
25             }else {
26                 stack.push(s.charAt(i));                
27             }
28         }
29         return stack.isEmpty();
30     }
31     
32     //方法一稍作改进版：将存储的括号字典改进，若遇到闭括号，但此时栈顶不为开括号，则直接返回false
33     public static boolean isValid11(String s) {
34         if(s.isEmpty()) {
35             return true;
36         }
37         if(s.length() == 1) {
38             return false;
39         }
40         Map<Character, Character> dict = new HashMap<>();
41         dict.put(‘)‘,‘(‘);
42         dict.put(‘]‘,‘[‘);
43         dict.put(‘}‘,‘{‘);
44         Stack<Character> stack = new Stack<>();
45         for(int i=0; i<s.length(); i++) {
46             if(dict.containsKey(s.charAt(i))) {
47                 char topChar = stack.isEmpty() ? ‘*‘ : stack.pop();
48                 if(dict.get(s.charAt(i)) != topChar) {
49                     return false;
50                 }
51             }else {
52                 stack.push(s.charAt(i));
53             }
54         }
55         return stack.isEmpty();
56     }
57     
58     //方法二：使用数组保存开括号，记录当前未匹配开括号的位置，若当前字符为闭括号，则与开括号比对
59     public boolean isValid2(String s) {
60         if(s == null || s.length() == 1) {
61             return false;
62         }
63         if(s.isEmpty()) {
64             return true;
65         }
66         Map<Character, Character> dict = new HashMap<>();
67         dict.put(‘)‘,‘(‘);
68         dict.put(‘]‘,‘[‘);
69         dict.put(‘}‘,‘{‘);
70         int len = s.length();
71         char[]    arr = new char[len];
72         int point = 0;
73         for(int i=0; i<len; i++) {
74             char c = s.charAt(i);
75             //若为开括号，直接保存入数组中
76             if(c == ‘(‘ || c == ‘[‘ || c == ‘{‘) {
77                 arr[point++] = c;
78             }
79             //若为闭括号，则与未匹配的开括号比对，若成功，则将未匹配开括号位置左移一位
80             else if(point > 0 && (arr[point-1] == dict.get(c))) {
81                 point--;
82             }else {
83                 return false;
84             }
85         }
86         return point == 0;
87     }