fix:最接近的三数之和

01-数组/day04/problem_solving_07.py

+"""
+最接近的三数之和
+"""
+import heapq
+from typing import List
+
+
+class Solution:
+    def threeSumClosest(self, nums: List[int], target: int) -> int:
+        """
+        思路:最关键的是找出最接近的数
+        :param nums:
+        :param target:
+        :return:
+        """
+        nums.sort()
+        ans = nums[0] + nums[1] + nums[2]
+        for i in range(len(nums)):
+            left = i + 1
+            right = len(nums) - 1
+            while left < right:
+                total = nums[i] + nums[left] + nums[right]
+                if abs(target - total) < abs(target - ans):
+                    ans = total
+                if total > target:
+                    right -= 1
+                elif total < target:
+                    left += 1
+                else:
+                    return total
+        return ans
+
+
+if __name__ == '__main__':
+    result = Solution().threeSumClosest([-1, 2, 1, -4], 1)
+    print(result)

02-链表/problem_solving_06.py

+"""
+相交链表
+"""
+from typing import List, Optional
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
+        """
+        思路:如果存在交点,则肯定在两个链表分别遍历对方时相交
+        :param headA:
+        :param headB:
+        :return:
+        """
+        if not headA or not headB:
+            return None
+        nodeA, nodeB = headA, headB
+        while nodeA != nodeB:
+            nodeA = nodeA.next if nodeA else headB
+            nodeB = nodeB.next if nodeB else headA
+        return nodeA
+
+
+if __name__ == '__main__':
+    l1 = ListNode(4)
+    l1.next = ListNode(1)
+    l1.next.next = ListNode(8)
+    l1.next.next.next = ListNode(4)
+    l1.next.next.next.next = ListNode(5)
+    # l1.next.next.next.next.next = ListNode(9)
+    # l1.next.next.next.next.next.next = ListNode(9)
+    l2 = ListNode(5)
+    l2.next = ListNode(6)
+    l2.next.next = ListNode(1)
+    l2.next.next.next = ListNode(8)
+    l2.next.next.next.next = ListNode(4)
+    l2.next.next.next.next.next = ListNode(5)
+    result = Solution().getIntersectionNode(l1, l2)
+    print(result)