*LeetCode--Valid Parentheses
Given a string containing just the characters '('
, ')'
, '{'
, '}'
, '['
and ']'
, determine if the input string is valid.
An input string is valid if:
- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input: "()" Output: true
Example 2:
Input: "()[]{}" Output: true
Example 3:
Input: "(]" Output: false
Example 4:
Input: "([)]" Output: false
Example 5:
Input: "{[]}" Output: true
自己的做法:因为不想一个一个分情况,所以考虑用map来进行数据的配对,然后用stack来进行符号的记录,
当前字符是左半边的符号时,入栈,
是右半边的符号时,进行对比,如果stack.peek()和它的另外一半相同,则出栈,否则返回false
最后,如果stack为空,则为true,否则为false
但是,事实证明,用map来装字符进行配对时的比较是错误的决定,这样会使得问题更加繁琐。
class Solution { public boolean isValid(String s) { if(s == null || s.length() == 0) return true; Stack<Character> stack = new Stack<>(); Map<Character, Character> map = new HashMap<>(); map.put('(', ')'); map.put('[', ']'); map.put('{', '}'); map.put(')', '('); map.put(']', '['); map.put('}', '{'); String str = "[{("; for(int i = 0; i < s.length(); i++){ char temp = s.charAt(i); if(map.get(temp) != null){ if(str.contains(String.valueOf(temp))){ stack.push(temp); } else{ if(!stack.empty() && stack.peek() == map.get(temp)){ stack.pop(); } else{ return false; } } } else{ return false; } } if(!stack.empty()) return false; return true; } }
discuss上的解法
只利用栈,然后将可能性列出来
public boolean isValid(String s) { Stack<Character> stack = new Stack<Character>(); for (char c : s.toCharArray()) { if (c == '(') stack.push(')'); else if (c == '{') stack.push('}'); else if (c == '[') stack.push(']'); else if (stack.isEmpty() || stack.pop() != c) return false; } return stack.isEmpty(); }
还可以利用replace()方法
public class Solution { public boolean isValid(String s) { int length; do { length = s.length(); s = s.replace("()", "").replace("{}", "").replace("[]", ""); } while(length != s.length()); return s.length() == 0; } }