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#yyds干货盘点# leetcode-栈-20


数据结构-栈,先进后出,匹配括号时,可以将左括号放到栈中,如果遇到合适的右括号就出栈,最后如果栈是空的就代表成功



import java.util.ArrayList;
import java.util.Stack;

/**
<p>给定一个只包括 <code>'('</code>,<code>')'</code>,<code>'{'</code>,<code>'}'</code>,<code>'['</code>,<code>']'</code> 的字符串 <code>s</code> ,判断字符串是否有效。</p>

<p>有效字符串需满足:</p>

<ol>
<li>左括号必须用相同类型的右括号闭合。</li>
<li>左括号必须以正确的顺序闭合。</li>
</ol>

<p> </p>

<p><strong>示例 1:</strong></p>

<pre>
<strong>输入:</strong>s = "()"
<strong>输出:</strong>true
</pre>

<p><strong>示例 2:</strong></p>

<pre>
<strong>输入:</strong>s = "()[]{}"
<strong>输出:</strong>true
</pre>

<p><strong>示例 3:</strong></p>

<pre>
<strong>输入:</strong>s = "(]"
<strong>输出:</strong>false
</pre>

<p><strong>示例 4:</strong></p>

<pre>
<strong>输入:</strong>s = "([)]"
<strong>输出:</strong>false
</pre>

<p><strong>示例 5:</strong></p>

<pre>
<strong>输入:</strong>s = "{[]}"
<strong>输出:</strong>true</pre>

<p> </p>

<p><strong>提示:</strong></p>

<ul>
<li><code>1 <= s.length <= 10<sup>4</sup></code></li>
<li><code>s</code> 仅由括号 <code>'()[]{}'</code> 组成</li>
</ul>
<div><div>Related Topics</div><div><li>栈</li><li>字符串</li></div></div><br><div><li>👍 3187</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
Character c = s.charAt(i);
if(c=='{'||c=='('||c=='['){
stack.push(c);
}else{
if(stack.isEmpty()){
return false;
}

if(!stack.isEmpty()){
if(stack.peek()=='{' && c=='}'){
stack.pop();
}else if(stack.peek()=='[' && c==']'){
stack.pop();
}else if(stack.peek()=='(' && c==')'){
stack.pop();
}else{
return false;
}
}
}
}
return stack.isEmpty();
}

}
//leetcode submit region end(Prohibit modification and deletion)


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