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复杂度计算_action

作者:互联网

 

有效的括号 - LeetCode 阅读
https://leetcode-cn.com/articles/valid-parentheses/

 

给定一个只包括 '('')''{''}''['']' 的字符串,判断字符串是否有效。

有效字符串需满足:

  1. 左括号必须用相同类型的右括号闭合。
  2. 左括号必须以正确的顺序闭合。

注意空字符串可被认为是有效字符串。

 

让我们看看使用栈作为该问题的中间数据结构的算法。

算法

  1. 初始化栈 S。
  2. 一次处理表达式的每个括号。
  3. 如果遇到开括号,我们只需将其推到栈上即可。这意味着我们将稍后处理它,让我们简单地转到前面的 子表达式。
  4. 如果我们遇到一个闭括号,那么我们检查栈顶的元素。如果栈顶的元素是一个 相同类型的 左括号,那么我们将它从栈中弹出并继续处理。否则,这意味着表达式无效。
  5. 如果到最后我们剩下的栈中仍然有元素,那么这意味着表达式无效。

 

 

 

class Solution {

// 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 Solution() {
this.mappings = new HashMap<Character, Character>();
this.mappings.put(')', '(');
this.mappings.put('}', '{');
this.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.empty() ? '#' : 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();
}
}

 

class Solution(object):
def isValid(self, s):
"""
:type s: str
:rtype: bool
"""

# The stack to keep track of opening brackets.
stack = []

# Hash map for keeping track of mappings. This keeps the code very clean.
# Also makes adding more types of parenthesis easier
mapping = {")": "(", "}": "{", "]": "["}

# For every bracket in the expression.
for char in s:

# If the character is an closing bracket
if char in mapping:

# Pop the topmost element from the stack, if it is non empty
# Otherwise assign a dummy value of '#' to the top_element variable
top_element = stack.pop() if stack else '#'

# The mapping for the opening bracket in our hash and the top
# element of the stack don't match, return False
if mapping[char] != top_element:
return False
else:
# We have an opening bracket, simply push it onto the stack.
stack.append(char)

# In the end, if the stack is empty, then we have a valid expression.
# The stack won't be empty for cases like ((()
return not stack

复杂度分析

 

标签:return,mappings,复杂度,element,括号,bracket,计算,action,stack
来源: https://www.cnblogs.com/yuanjiangw/p/10798429.html