238. Product of Array Except Self -M

Given an array nums of n integers where n > 1,  return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]

Note: Please solve it without division and in O(n).

Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

A:

从前向后走一遍，计算之前的乘积
再从后向前走一遍，计算之后的乘积，

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n,1);
        // two round, forward, and backward
        for(int i =1;i<n;++i)
            res[i] = res[i-1] * nums[i-1];
        int back = 1;
        for(int i = n-1; i>=0; --i)
        {
            res[i] *= back;
            back *= nums[i];
        }
        return res;
    }
};

用2个帮助数组

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int pre[] = new int[n];
        int after[] = new int[n];
        pre[0] = 1;
        for(int i =1;i<n;i++)
            pre[i] = pre[i-1]*nums[i-1];
        after[n-1] = 1;
        for(int i=n-2;i>=0;i--)
            after[i] = after[i+1] * nums[i+1];
        int[] result = new int[n];
        for(int i =0;i<n;i++)
            result[i] = pre[i] * after[i];
        return result;
    }
}

Follow up:
为节省空间，用result数组和给的nums数组代替pre   after 数组。

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] pre = new int[n];
        pre[0] = 1;
        for(int i =1;i<n;i++)
            pre[i] = pre[i-1]*nums[i-1];
        for(int i=n-2;i>0;i--)
            nums[i] = nums[i+1] * nums[i];
        for(int i =0;i<n-1;i++)
            pre[i] = pre[i] * nums[i+1];
        return pre;
    }
}

Mistakes:

237. Delete Node in a Linked List (M)

Q:

There is a singly-linked list head and we want to delete a node node in it.

You are given the node to be deleted node. You will not be given access to the first node of head.

All the values of the linked list are unique, and it is guaranteed that the given node node is not the last node in the linked list.

Delete the given node. Note that by deleting the node, we do not mean removing it from memory. We mean:

• The value of the given node should not exist in the linked list.
• The number of nodes in the linked list should decrease by one.
• All the values before node should be in the same order.
• All the values after node should be in the same order.

Custom testing:

• For the input, you should provide the entire linked list head and the node to be given node. node should not be the last node of the list and should be an actual node in the list.
• We will build the linked list and pass the node to your function.
• The output will be the entire list after calling your function.

 

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

 

Constraints:

• The number of the nodes in the given list is in the range [2, 1000].
• -1000 <= Node.val <= 1000
• The value of each node in the list is unique.
• The node to be deleted is in the list and is not a tail node.

A:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        ListNode* pre = node;
        while(pre->next != NULL && pre->next->next != NULL){
            pre->val = pre->next->val;
            pre = pre->next;
        }
        pre->val = pre->next->val;
        pre->next = NULL;
    }
};

Mistakes:
最新一次代码，好几年不写了。
应该update node的value before node move next

236. Lowest Common Ancestor of a Binary Tree. (M)

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

 

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

A:


Google 面试题   哎，惨痛～～～～～～～～～～

/**
 * 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);
        if(L&&R)
            return root;
        return L?L:R;
    }
};

Mistakes:

235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

 

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [2,1], p = 2, q = 1
Output: 2

 

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

A:
根据val ， 看左边右边

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(p.val > q.val)
            return lowestCommonAncestor(root,q,p);
        return helper(root,p,q);
    }
    private TreeNode helper(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val<p.val)
            return helper(root.right,p,q);
        else if(root.val > q.val)
            return helper(root.left,p,q);
        return root;
    }
}

其实这道题，无关Binary Search Tree ---虽然利用起来，效率会高一些些

/**
 * 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)
            return root;
        if(root->val == p->val || root->val == q->val){
            return root;
        }
        auto l = lowestCommonAncestor(root->left, p, q);
        auto r = lowestCommonAncestor(root->right, p, q);
        if(l&&r){
            return root;
        }else{
            return l?l:r;
        }
    }
};

Mistakes: