【LeetCode 678】 Valid Parenthesis String

题目描述

Given a string containing only three types of characters: ‘(’, ‘)’ and ‘*’, write a function to check whether this string is valid. We define the validity of a string by these rules:

Any left parenthesis ‘(’ must have a corresponding right parenthesis ‘)’.
Any right parenthesis ‘)’ must have a corresponding left parenthesis ‘(’.
Left parenthesis ‘(’ must go before the corresponding right parenthesis ‘)’.
‘*’ could be treated as a single right parenthesis ‘)’ or a single left parenthesis ‘(’ or an empty string.
An empty string is also valid.

Example 1:

Input: "()"
Output: True

Example 2:

Input: "(*)"
Output: True

Example 3:

Input: "(*))"
Output: True

Note:

The string size will be in the range [1, 100].

思路

1. 动态规划：O （n^3） [i][j]范围是否合法，可以从[i-1][j-1] [i][j-1] [i-1][j-1] 推出。还有可能是从某个点 k ，[i][k] [k+1][j] 两个得到的。
2. 栈：左括号栈和右括号栈，遇见左括号或者星号，入栈。遇见右括号，用左括号抵消或者星号抵消。最后用星号抵消左括号，条件是星号在左括号右边。时间复杂度：O (n)

代码

代码一：

class Solution {
public:
    bool check(char a, char b) {
         if ((a == '(' && b == '*') || (a == '*' && b == ')') || (a == '*' && b == '*') || (a == '(' && b == ')'))
             return true;
        return false;
    }
    
    bool checkValidString(string s) {
        if (s.empty()) return true;
        int n = s.length();
        vector<vector<int> > dp(n, vector<int>(n, 0));
        
        for (int l=1; l<=n; ++l) {
            for (int i=0, j=i+l-1; j<n; ++i, ++j) {
                if (l == 1) {
                    if (s[i] == '*')
                        dp[i][j] = 1;
                }
                else if (j == i+1) {
                    if (check(s[i], s[j])) dp[i][j] = 1;
                }
                else {
                    if ((dp[i+1][j-1] && check(s[i], s[j])) || (dp[i][j-1] && s[j] == '*') || (dp[i+1][j] && s[i] == '*'))
                        dp[i][j] = 1;
                }
                for (int k=i; k<j; ++k) {
                    dp[i][j] |= (dp[i][k] && dp[k+1][j]);
                }
            }
        }
        
        return dp[0][n-1];
    }
};

代码二：

class Solution {
public:
    bool checkValidString(string s) {
        stack<int> left;
        stack<int> star;
        int n = s.length();
        for (int i=0; i<n; ++i) {
            if (s[i] == '(') left.push(i);
            else if (s[i] == '*') star.push(i);
            else {
                if (!left.empty()) left.pop();
                else if (!star.empty()) star.pop();
                else return false;
            }
        }
        
        while(!left.empty() && !star.empty()) {
            if (star.top() < left.top()) return false;
            left.pop();
            star.pop();
        }
        
        return left.empty();
    }
};