版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/m0_38088298/article/details/85854314
- 描述：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.
Example 1:
Input: “()”
Output: true
Example 2:
Input: “()[]{}”
Output: true
Example 3:
Input: “(]”
Output: false
Example 4:
Input: “([)]”
Output: false
Example 5:
Input: “{[]}”
Output: true
- 分析：给出一个表达式只含有‘(’‘[’‘{’三种括号，要求判断此表达式是否符合括号匹配规则：
1.括号类型匹配
2.括号顺序匹配 - 思路一：使用栈来解决。如果遇到‘(’‘[’‘{’三种开括号，就将它们push进栈，遇到‘)’‘]’‘}’三种开括号，判断栈顶元素是否是与之匹配的开括号，是则将栈顶元素pop出去，不是则return false。
class Solution {
public:
bool isValid(string s) {
stack<char> character_match;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '(' || s[i] == '[' || s[i] == '{') {
character_match.push(s[i]);
}
if (s[i] == ')') {
if (!character_match.empty() && character_match.top() == '(') character_match.pop();
else return false;
}
if (s[i] == ']') {
if (!character_match.empty() && character_match.top() == '[') character_match.pop();
else return false;
}
if (s[i] == '}') {
if (!character_match.empty() && character_match.top() == '{') character_match.pop();
else return false;
}
}
return character_match.empty();
}
};
- 思路二：用map优化一下代码。
class Solution {
public:
bool isValid(string s) {
stack<char> character_match;
map<char, char> char_mapping;
char_mapping.insert(make_pair(')', '('));
char_mapping.insert(make_pair(']', '['));
char_mapping.insert(make_pair('}', '{'));
for (int i = 0; i < s.size(); i++) {
if (s[i] == '(' || s[i] == '[' || s[i] == '{') character_match.push(s[i]);
else if (!character_match.empty() && char_mapping.find(s[i]) != char_mapping.end() && character_match.top() == char_mapping[s[i]]) character_match.pop();
else return false;
}
return character_match.empty();
}
};