Leetcode-字符串-20

20. Valid Parentheses

题目链接

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.

Input: ” {[]} ” Output: true
Input: ” ([)] ” Output: false
Input: ” ()[]{} ” Output: true

判断多种括号(){}[]是否完整配对

方法1：5ms

使用java.util.Stack栈
每次读到前括号“ ( [ { ”时，不压入该前括号，而是向栈中压入其对应的后括号“ ) ] } ”。
这样在之后读到后括号时，就无需再额外找对应的前括号来判断是否与栈顶匹配，而是直接判断该后括号与栈顶是否相等。

public boolean isValid(String s) {
        Stack<Character> stack = new Stack<Character>();
        for (char ch : s.toCharArray()){
            if (ch == '(')
                stack.push(')');
            else if (ch =='[')
                stack.push(']');
            else if (ch =='{')
                stack.push('}');
            else if (stack.isEmpty() || stack.pop() != ch)
                return false;
        }
        return stack.isEmpty();
}

方法2：5ms

使用java.util.Stack栈
与方法1不同的是，每次读到前括号“ ( [ { ”时，直接压入该前括号

public boolean isValid3(String s) {
        Stack<Character> stack = new Stack<Character>();
        String str1 = "({[";
        String str2 = ")}]";
        for (char ch : s.toCharArray()){
            if (str1.indexOf(ch) != -1)
                stack.push(ch);
            else if (str2.indexOf(ch) != -1){
                if (stack.isEmpty() || stack.peek() != str1.charAt(str2.indexOf(ch)))
                    return false;
                else
                    stack.pop();
            }
          //else if (stack.isEmpty()||stack.pop() != str1.charAt(str2.indexOf(ch))) return false;

        }
        return stack.isEmpty();
    }

方法3：4ms

没直接使用java的java.util.Stack栈。而是用数组来模拟栈操作

public boolean isValid2(String s) {
        char[] stack = new char[s.length()];
        String str1 = "({[";
        String str2 = ")}]";
        int j = 0;
        for (int i = 0; i < s.length(); i++) {
            if (str1.indexOf(s.charAt(i)) != -1){
                stack[j++] = s.charAt(i);
            }
            else if (str2.indexOf(s.charAt(i)) != -1){
                if (j == 0 || stack[j - 1] != str1.charAt(str2.indexOf(s.charAt(i))))
                    return false;
                else
                    j = j - 1;
            }
        }
        return j == 0;
    }