feat: java for $20

azl397985856 · web-flow · commit 7736544bd733 · 2022-09-23T13:24:58.000+08:00
diff --git a/problems/20.valid-parentheses.md b/problems/20.valid-parentheses.md
@@ -88,7 +88,7 @@ https://leetcode-cn.com/problems/valid-parentheses/description
 
 ### 代码
 
-代码支持：JS，Python
+代码支持：JS，Python，Java，C++
 
 Javascript Code:
 
@@ -150,6 +150,66 @@ Python Code:
           return len(stack) == 0
 ```
 
+Java Code:
+
+```java
+class Solution {
+    public boolean isValid(String s) {
+        //1.判断空字符串
+        if(s.isEmpty()) return true;
+        //2.创建辅助栈
+        Stack<Character> stack = new Stack<>();
+        //3.仅遍历一次
+        for(char c : s.toCharArray()){
+            if(c == '('){
+                stack.push(')');
+            }else if(c == '['){
+                stack.push(']');
+            }else if(c == '{'){
+                stack.push('}');
+            }else if(stack.isEmpty() || c != stack.pop()){
+                return false;
+            }
+        }
+        //4.返回
+        return stack.isEmpty();
+    }
+}
+```
+
+C++ Code:
+
+```cpp
+class Solution {
+public:
+    bool isValid(string s) {
+        int n = s.size();
+        if (n % 2 == 1) {
+            return false;
+        }
+
+        unordered_map<char, char> pairs = {
+            {')', '('},
+            {']', '['},
+            {'}', '{'}
+        };
+        stack<char> stk;
+        for (char ch: s) {
+            if (pairs.count(ch)) {
+                if (stk.empty() || stk.top() != pairs[ch]) {
+                    return false;
+                }
+                stk.pop();
+            }
+            else {
+                stk.push(ch);
+            }
+        }
+        return stk.empty();
+    }
+};
+```
+
 **复杂度分析**
 
 - 时间复杂度：$O(N)$
