字符串中的匹配之递归

字符串的括号匹配是一个很常见的问题。用栈这种后进先出的结构是非常适合的。此外，字符串中的回文以及衍生的各种问题也是字符串处理中非常常见的。

今天再说一下这类相似的问题，如何用递归来转化成子结构来求解。

先放一条LeetCode例题: 680. Valid Palindrome II

Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
Example 1:
Input: "aba"
Output: True

Example 2:
Input: "abca"
Output: True

题目的意思是：至多删除一个字母判断是否能删成回文。字符串长度是50000.

显然不能O(n^2)暴力。然后我们分析一下，其实我们只要能找出这个有问题的字母的位置在哪里就好了，然后删掉它就行了。可以通过判断是哪个字母的个数是奇数来做吗？不能，因为判断出来是什么字母，但当这个字母有很多时我们还是不知道删掉其中的哪一个。

所以就用到所说的递归:

两个指针从左边和右分别扫，扫到不匹配的时候就递归，分别删掉左边或右边的这个字母来判断新串是不是回文。

注意的是，已经判断回文的部分就不需要递归了，否则就起不到递归的作用了。时间复杂度: O(n)

AC代码：

class Solution {
public:
    bool isHuiwen(string& s,int l,int r){
        while (l < r){
            if (s[l] != s[r])
                return false;
            l++;
            r--;
        }
        return true;
    }
    bool validPalindrome(string s) {
        bool res;
        int l, r;
        l = 0;
        r = s.size()-1;
        while (l < r){
            if (s[l] != s[r])
                return isHuiwen(s, l+1, r) || isHuiwen(s, l, r-1);
            l++;
            r--;
        }
        return isHuiwen(s, 0, s.size()-1);
    }
};

再放一道LeetCode例题: 678. Valid Parenthesis String

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:

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

Example 1:
Input: "()"
Output: True
Example 2:
Input: "(*)"
Output: True
Example 3:
Input: "(*))"
Output: True

题意就是判断是否括号匹配，其中*可任意充当(或者)或者空白。

如果是没有*，那么做法很显然：从左到右统计(和)的个数。如果)的个数>（的个数就不行。判断到结尾的时候，如果两者个数相等就可以。

现在引入了*这个百搭，我们同样可以用递归来做：

如果遇到一个*，就分别把它视为左括号或者右括号或者空白来对后面的子串递归。

AC代码:

class Solution2 {
public:
    int lll;

    bool check(string& s,int start,int lcnt,int rcnt){
        int l, r;
        l = lcnt;
        r = rcnt;
        for (int i = start; i < lll; ++i){
            if (s[i] == ‘(‘)
                l++;
            if (s[i] == ‘)‘)
                r++;
            if (r > l)
                return false;
            if (s[i]==‘*‘){
                return check(s, i+1, l+1, r) ||
                       check(s, i+1, l, r+1) ||
                        check(s, i+1, l, r);
            }
        }
        return l == r;
    }

    bool checkValidString(string s) {
        lll = s.size();
        return check(s, 0, 0, 0);
    }
};

另外，这道题再说一个很巧妙的做法，思路来源于讨论区的高票帖。

low代表左括号可能出现的下界，high代表左括号可能出现的上界。

  • 当遇到一个(时，low++,high++;
  • 当遇到一个)时，low--,high--;
  • 当遇到一个*时，low--,high++; （即我们既可以把*看成(也可以看成)
  • 注意的是，下界一旦小于0直接返回false代表右括号太多了， 但是上界我们得人为保证其值>=0，因为*可以被视作空白或右括号(例如"(**")。但是最后结束的时候必须保证上界为0，如果最后上界不为0就是左括号太多了。

另附AC代码:

class Solution {
public:
    bool checkValidString(string s) {
        int lower = 0, upper = 0;
        for (char c : s) {
            if (c==‘(‘) {
                lower++;
                upper++;
            } else if (c==‘)‘) {
                lower--;
                upper--;
            } else { // * encountered
                lower--;
                upper++;
            }
            lower = max(lower, 0);
            if (upper<0) // unmatched ‘)‘ found in the middle of string
                return false;
        }
        return lower==0;
    }
};

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时间： 2024-12-20 09:20:38

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