LeetCode-20.Valid Parentheses

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. 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

public boolean isValid(String s) {//栈 my
        Stack<Character> stack = new Stack<Character>();
        for (int i = 0; i <s.length() ; i++) {
            char c = s.charAt(i);
            if('('==c||'['==c||'{'==c){
                stack.push(c);
            }
            else {
                if(stack.empty()){
                    return false;
                }
                Character top = stack.pop();
                if((')'==c&&'('!=top)||(']'==c&&'['!=top)||('}'==c&&'{'!=top)) {
                    return false;
                }
            }
        }
        if(!stack.empty()){
            return false;
        }
        return true;
    }　　

可扩展版

class Solution {
 
  // Hash table that takes care of the mappings.
  private HashMap<Character, Character> mappings;
 
  // Initialize hash map with mappings. This simply makes the code easier to read.
  public Solution() {
    this.mappings = new HashMap<Character, Character>();
    this.mappings.put(')', '(');
    this.mappings.put('}', '{');
    this.mappings.put(']', '[');
  }
 
  public boolean isValid(String s) {
 
    // Initialize a stack to be used in the algorithm.
    Stack<Character> stack = new Stack<Character>();
 
    for (int i = 0; i < s.length(); i++) {
      char c = s.charAt(i);
 
      // If the current character is a closing bracket.
      if (this.mappings.containsKey(c)) {
 
        // Get the top element of the stack. If the stack is empty, set a dummy value of '#'
        char topElement = stack.empty() ? '#' : stack.pop();
 
        // If the mapping for this bracket doesn't match the stack's top element, return false.
        if (topElement != this.mappings.get(c)) {
          return false;
        }
      } else {
        // If it was an opening bracket, push to the stack.
        stack.push(c);
      }
    }
 
    // If the stack still contains elements, then it is an invalid expression.
    return stack.isEmpty();
  }
}

　　

进阶题

生成有效括号组合 LeetCode22 https://www.cnblogs.com/zhacai/p/10599156.html