Description
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.
Examples
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
解题思路
就是堆栈的思想
一开始以为还需要判断嵌套关系，比如大括号一定要在最外层之类，但其实没用这个要求，用test测试了下 " ( { } ) " 这种也是返回true
那用堆栈就完事儿了
- 如果碰到了 { 、[ 、(，就push到堆栈中
- 如果碰到了 }、]、)，就判断堆栈的top是不是与之对应的符号
代码如下：
class Solution {
public boolean isValid(String s) {
Stack<Character> st = new Stack<>();
int i;
Set<Character> left = new HashSet<>();
left.add('(');
left.add('[');
left.add('{');
Map<Character, Character> k = new HashMap<>();
k.put(')', '(');
k.put(']', '[');
k.put('}', '{');
for(i = 0; i < s.length(); i++){
char t = s.charAt(i);
if(left.contains(t)){
st.push(t);
}
else{
if(st.empty())
return false;
if(st.peek() == k.get(t))
st.pop();
else
return false;
}
}
if(st.empty())
return true;
return false;
}
}
其他
- 参加了美赛学会了Excel hmmmmmm
- 保U冲O！
- 寒假start！