feat: add solutions to lc problem: No.1290. Convert Binary Number in a Linked List to Integer

Author: yanglbme

solution/1200-1299/1290.Convert Binary Number in a Linked List to Integer/README.md

@@ -55,27 +55,69 @@
 	<li>每个结点的值不是&nbsp;<code>0</code> 就是 <code>1</code>。</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+先求链表长度 n，再遍历链表，累加求得结果。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getDecimalValue(self, head: ListNode) -> int:
+        n, cur = 0, head
+        while cur:
+            n += 1
+            cur = cur.next
+        res, cur = 0, head
+        while cur:
+            n -= 1
+            if cur.val == 1:
+                res += (2 ** (n))
+            cur = cur.next
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public int getDecimalValue(ListNode head) {
+        int n = 0;
+        for (ListNode cur = head; cur != null; cur = cur.next) {
+            ++n;
+        }
+        int res = 0;
+        for (ListNode cur = head; cur != null; cur = cur.next) {
+            --n;
+            if (cur.val == 1) {
+                res += Math.pow(2, n);
+            }
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/1200-1299/1290.Convert Binary Number in a Linked List to Integer/README_EN.md

@@ -6,12 +6,8 @@
 
 <p>Given <code>head</code> which is a reference node to&nbsp;a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.</p>
 
-
-
 <p>Return the <em>decimal value</em> of the number in the linked list.</p>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Example 1:</strong></p>
@@ -28,12 +24,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 2:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> head = [0]
@@ -42,12 +34,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 3:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> head = [1]
@@ -56,12 +44,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 4:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0]
@@ -70,12 +54,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 5:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> head = [0,0]
@@ -84,14 +64,10 @@
 
 </pre>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Constraints:</strong></p>
 
-
-
 <ul>
 	<li>The Linked List is not empty.</li>
 	<li>Number of nodes&nbsp;will not exceed <code>30</code>.</li>
@@ -105,13 +81,54 @@
 ### **Python3**
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getDecimalValue(self, head: ListNode) -> int:
+        n, cur = 0, head
+        while cur:
+            n += 1
+            cur = cur.next
+        res, cur = 0, head
+        while cur:
+            n -= 1
+            if cur.val == 1:
+                res += (2 ** (n))
+            cur = cur.next
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public int getDecimalValue(ListNode head) {
+        int n = 0;
+        for (ListNode cur = head; cur != null; cur = cur.next) {
+            ++n;
+        }
+        int res = 0;
+        for (ListNode cur = head; cur != null; cur = cur.next) {
+            --n;
+            if (cur.val == 1) {
+                res += Math.pow(2, n);
+            }
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/1200-1299/1290.Convert Binary Number in a Linked List to Integer/Solution.java

@@ -8,15 +8,17 @@
  */
 class Solution {
     public int getDecimalValue(ListNode head) {
-        int sum = 0;
-        StringBuilder sb = new StringBuilder("0");
-        while (head != null) {
-            sum += head.val;
-            if (sum != 0) {
-                sb.append(head.val);
+        int n = 0;
+        for (ListNode cur = head; cur != null; cur = cur.next) {
+            ++n;
+        }
+        int res = 0;
+        for (ListNode cur = head; cur != null; cur = cur.next) {
+            --n;
+            if (cur.val == 1) {
+                res += Math.pow(2, n);
             }
-            head = head.next;
         }
-        return Integer.valueOf(sb.toString(), 2);
+        return res;
     }
 }

solution/1200-1299/1290.Convert Binary Number in a Linked List to Integer/Solution.py

@@ -0,0 +1,19 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getDecimalValue(self, head: ListNode) -> int:
+        n, cur = 0, head
+        while cur:
+            n += 1
+            cur = cur.next
+        res, cur = 0, head
+        while cur:
+            n -= 1
+            if cur.val == 1:
+                res += (2 ** (n))
+            cur = cur.next
+        return res