二叉搜索树的最近公共祖先
假如p, q在root同一侧，则继续递归这一侧，否则返回当前节点（可能当前节点为p,q之一，也可能在不同侧）
如果是BST,可以保存到p, q的路径，然后找到链表最后一个公共节点即可
235. Lowest Common Ancestor of a Binary Search Tree
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(p->val < root->val && q->val < root->val) return lowestCommonAncestor(root->left, p, q);
if(p->val > root->val && q->val > root->val) return lowestCommonAncestor(root->right, p, q);
return root;
}
};二叉树的最近公共祖先
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root || root == p || root == q) return root;
auto l = lowestCommonAncestor(root->left, p, q);
auto r = lowestCommonAncestor(root->right, p, q);
return !l ? r :!r ? l : root;
}
}两个链表的第一个公共节点
换着遍历，当遍历到长度为da（a独有） + db（b独有） + ab（ab公共部分）时，两指针会合，
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
auto p = headA;
auto q = headB;
while(p != q){
p = p ? p->next : headB;
q = q ? q->next : headA;
}
return p;
}
};还有一种方法是计算链表长度的不谈了，我当时的做法时用异或+反转链表, 异或的结果就是第一个交点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverse(ListNode* h, intptr_t & pivot){
auto cur = h;
ListNode* prev = NULL;
while(cur != NULL){
pivot = pivot ^ (intptr_t)(cur);
auto t = cur->next;
cur->next = prev;
prev = cur;
cur = t;
}
return prev;
}
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
intptr_t pivot = 0;
auto end = reverse(headA, pivot);
auto b = reverse(headB, pivot);
auto a = reverse(end, pivot);
if(b == headA){
return (ListNode*)pivot;
} else {
reverse(b, pivot);
return NULL;
}
}
};