算法leetcode｜32. 最长有效括号（rust重拳出击）

Posted 2023-01-20 二当家的白帽子

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文章目录

• 32. 最长有效括号：
• • 样例 1：
  • 样例 2：
  • 样例 3：
  • 提示：
• 分析：
• 题解：
• • rust
  • go
  • c++
  • c
  • python
  • java

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32. 最长有效括号：

给你一个只包含 '(' 和 ')' 的字符串，找出最长有效（格式正确且连续）括号子串的长度。

样例 1：

输入：
	s = "(()"
	
输出：
	2
	
解释：
	最长有效括号子串是 "()"

样例 2：

输入：
	s = ")()())"
	
输出：
	4
	
解释：
	最长有效括号子串是 "()()"

样例 3：

输入：
	s = ""
	
输出：
	0

提示：

• 0 <= s.length <= 3 * 10⁴
• s[i] 为 '(' 或 ')'

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分析：

• 面对这道算法题目，二当家的陷入了沉思。
• 只有小括号，判断有效是关键。
• 从左向右看，左括号数始终大于等于右括号数，并且最终左右括号数相等。
• 从右向左看，右括号数始终大于等于左括号数，并且最终左右括号数相等。

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题解：

rust

impl Solution 
    pub fn longest_valid_parentheses(s: String) -> i32 
        let (mut left, mut right, mut maxlength) = (0, 0, 0);
        s.as_bytes().iter().for_each(|&b| 
            if b == b'(' 
                left += 1;
             else 
                right += 1;
            
            if left == right 
                maxlength = maxlength.max(2 * right);
             else if right > left 
                left = 0;
                right = 0;
            
        );
        left = 0;
        right = 0;
        s.as_bytes().iter().rev().for_each(|&b| 
            if b == b'(' 
                left += 1;
             else 
                right += 1;
            
            if left == right 
                maxlength = maxlength.max(2 * left);
             else if left > right 
                left = 0;
                right = 0;
            
        );
        return maxlength;
    

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go

func longestValidParentheses(s string) int 
    max := func(x, y int) int 
		if x > y 
			return x
		
		return y
	

	left, right, maxLength := 0, 0, 0
	for i := 0; i < len(s); i++ 
		if s[i] == '(' 
			left++
		 else 
			right++
		
		if left == right 
			maxLength = max(maxLength, 2*right)
		 else if right > left 
			left, right = 0, 0
		
	
	left, right = 0, 0
	for i := len(s) - 1; i >= 0; i-- 
		if s[i] == '(' 
			left++
		 else 
			right++
		
		if left == right 
			maxLength = max(maxLength, 2*left)
		 else if left > right 
			left, right = 0, 0
		
	
	return maxLength

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c++

class Solution 
public:
    int longestValidParentheses(string s) 
        int left = 0, right = 0, maxlength = 0;
        for (int i = 0; i < s.length(); ++i) 
            if (s[i] == '(') 
                left++;
             else 
                right++;
            
            if (left == right) 
                maxlength = max(maxlength, 2 * right);
             else if (right > left) 
                left = right = 0;
            
        
        left = right = 0;
        for (int i = (int) s.length() - 1; i >= 0; --i) 
            if (s[i] == '(') 
                left++;
             else 
                right++;
            
            if (left == right) 
                maxlength = max(maxlength, 2 * left);
             else if (left > right) 
                left = right = 0;
            
        
        return maxlength;
    
;

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c

int longestValidParentheses(char * s)
    const int len = strlen(s);
    int left = 0, right = 0, maxlength = 0;
    for (int i = 0; i < len; ++i) 
        if (s[i] == '(') 
            left++;
         else 
            right++;
        
        if (left == right) 
            maxlength = fmax(maxlength, 2 * right);
         else if (right > left) 
            left = right = 0;
        
    
    left = right = 0;
    for (int i = len - 1; i >= 0; --i) 
        if (s[i] == '(') 
            left++;
         else 
            right++;
        
        if (left == right) 
            maxlength = fmax(maxlength, 2 * left);
         else if (left > right) 
            left = right = 0;
        
    
    return maxlength;

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python

class Solution:
    def longestValidParentheses(self, s: str) -> int:
        left, right, max_length = 0, 0, 0
        for i in range(len(s)):
            if s[i] == '(':
                left += 1
            else:
                right += 1
            if left == right:
                max_length = max(max_length, 2 * right)
            elif right > left:
                left, right = 0, 0
        left, right = 0, 0
        for i in range(len(s) - 1, -1, -1):
            if s[i] == '(':
                left += 1
            else:
                right += 1
            if left == right:
                max_length = max(max_length, 2 * left)
            elif left > right:
                left, right = 0, 0
        return max_length

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java

class Solution 
    public int longestValidParentheses(String s) 
        int left = 0, right = 0, maxlength = 0;
        for (int i = 0; i < s.length(); ++i) 
            if (s.charAt(i) == '(') 
                left++;
             else 
                right++;
            
            if (left == right) 
                maxlength = Math.max(maxlength, 2 * right);
             else if (right > left) 
                left = right = 0;
            
        
        left = right = 0;
        for (int i = s.length() - 1; i >= 0; --i) 
            if (s.charAt(i) == '(') 
                left++;
             else 
                right++;
            
            if (left == right) 
                maxlength = Math.max(maxlength, 2 * left);
             else if (left > right) 
                left = right = 0;
            
        
        return maxlength;
    

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