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  • LeetCode_20. Valid Parentheses

    20. Valid Parentheses

    Easy

    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
    package leetcode.easy;
    
    import java.util.HashMap;
    import java.util.Stack;
    
    public class ValidParentheses {
    
    	// 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 ValidParentheses() {
    		mappings = new HashMap<Character, Character>();
    		mappings.put(')', '(');
    		mappings.put('}', '{');
    		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.isEmpty() ? '#' : 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();
    	}
    
    	@org.junit.Test
    	public void test() {
    		String s1 = "()";
    		String s2 = "()[]{}";
    		String s3 = "(]";
    		String s4 = "([)]";
    		String s5 = "{[]}";
    
    		System.out.println(isValid(s1));
    		System.out.println(isValid(s2));
    		System.out.println(isValid(s3));
    		System.out.println(isValid(s4));
    		System.out.println(isValid(s5));
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/denggelin/p/11535065.html
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