链表中倒数第k个节点

CodingProgrammer · web-flow · commit d7501269f0bd · 2020-11-20T13:47:07.000+08:00
diff --git a/双指针/剑指 Offer 22.py b/双指针/剑指 Offer 22.py
@@ -0,0 +1,89 @@
+'''
+Descripttion: 链表中倒数第k个节点
+version: 1
+Author: Jason
+Date: 2020-11-20 12:08:11
+LastEditors: Jason
+LastEditTime: 2020-11-20 13:42:23
+'''
+# Definition for singly-linked list.
+import random
+
+
+def GenerateRandomList(number, size):
+    temp = list()
+    random_legth = random.randint(0, size)
+    current_length = 0
+    while current_length < random_legth:
+        temp.append(random.randint(1, number))
+        current_length += 1
+    return temp
+
+
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+
+class Solution:
+    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
+        if not head:
+            return None
+        fast = head
+        slow = head
+        while k > 0:
+            fast = fast.next
+            k -= 1
+
+        while fast:
+            fast = fast.next
+            slow = slow.next
+        return slow
+
+    def getKthFromEnd2(self, head: ListNode, k: int) -> ListNode:
+        if not head:
+            return None
+        # 首先算出链表的总长度
+        cur = head
+        length = 0
+        while cur:
+            length += 1
+            cur = cur.next
+
+        cur = head
+        times = length - k
+        while times > 0:
+            cur = cur.next
+            times -= 1
+
+        return cur
+
+
+if __name__ == "__main__":
+    s = Solution()
+    for _ in range(100):
+        nums = GenerateRandomList(20, 10)
+        if len(nums) < 1:
+            continue
+        nodes = [ListNode(each_num) for each_num in nums]
+        # 连接链表
+        for i in range(len(nodes) - 1):
+            nodes[i].next = nodes[i + 1]
+        k = random.randint(1, len(nums))
+        if s.getKthFromEnd(nodes[0], k).val != nums[-k]:
+            print("Wrong:", nums)
+    print("Done!")
+    '''
+    n1 = ListNode(1)
+    n2 = ListNode(2)
+    n3 = ListNode(3)
+    n4 = ListNode(4)
+    n5 = ListNode(5)
+    n1.next = n2
+    n2.next = n3
+    n3.next = n4
+    n4.next = n5
+
+    print(s.getKthFromEnd(n1, 3).val)
+    '''
