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Best Time to Buy and Sell Stock II

 Total Accepted: 22240 Total Submissions: 60488My Submissions

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

class Solution {public:    int maxProfit(vector<int> &prices)     {        if(prices.size() < 2)            return 0;        int profit = 0;        for(int i = 1; i < prices.size();i++)        {            if(prices[i] > prices[i - 1])                profit += prices[i] - prices[i - 1];                    }        return profit;    }};

2.Unique Binary Search Trees

 Total Accepted: 23078 Total Submissions: 63359My Submissions

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.

   1         3     3      2      1    \       /     /      / \      \     3     2     1      1   3      2    /     /       \                 \   2     1         2                 3

//参考别人的class Solution {public:    int numTrees(int n)     {          vector<int> v(n+1,0);          v[0] = 1;          v[1] = 1;          v[2] = 2;          if(n <= 1)          return 1;          else if(n == 2)          return 2;          for(int i = 3;i <= n; i++)          {              for(int j = 1; j <= i; j++)              {                  v[i] += v[j - 1] * v[i - j];              }          }          return v[n];              }};

3.Linked List Cycle

 Total Accepted: 29113 Total Submissions: 81160My Submissions

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    bool hasCycle(ListNode *head)     {        ListNode *p = head,*q = head;        while(q!= NULL && q->next != NULL)        {            p = p->next;             q = q->next->next;            if(p == q)            return true;                    }        return false;            }};

4.Binary Tree Inorder Traversal

 Total Accepted: 29287 Total Submissions: 82207My Submissions

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

class Solution {public:void inorder(TreeNode *root,vector<int>&result){if(root == NULL)return;inorder(root->left,result);result.push_back(root->val);inorder(root->right,result);}    vector<int> inorderTraversal(TreeNode *root) {vector<int> result;inorder(root,result);return result;            }};

5.Binary Tree Preorder Traversal

 Total Accepted: 32064 Total Submissions: 90031My Submissions

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {        vector<int>result;        result.clear();        stack<TreeNode*>st;        if(root == NULL)            return result;        st.push(root);        while(!st.empty())        {            TreeNode *node = st.top();            st.pop();            result.push_back(node->val);            if(NULL != node->right)            {                    st.push(node->right);                            }            if(NULL != node->left)            {                st.push(node->left);            }        }return result;    }};

class Solution {public:void preorder(TreeNode *root,vector<int> &result){if(root == NULL)return;result.push_back(root->val);preorder(root->left,result);preorder(root->right,result);}    vector<int> preorderTraversal(TreeNode *root) {        vector<int>result;preorder(root,result);return result;    }};

6.Populating Next Right Pointers in Each Node

 Total Accepted: 23205 Total Submissions: 65706My Submissions

Given a binary tree

    struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1       /  \      2    3     / \  / \    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \  / \    4->5->6->7 -> NULL

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { *  int val; *  TreeLinkNode *left, *right, *next; *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public:    void connect(TreeLinkNode *root)    {        if(root == NULL)            return;        TreeLinkNode *left = root->left;        TreeLinkNode *right = root->right;        if(NULL != left)        {            left->next = right;        }        if(NULL !=right && NULL !=root->next)        {            right->next = root->next->left;        }        connect(left);        connect(right);            }};

7.Search Insert Position

 Total Accepted: 24348 Total Submissions: 69507My Submissions

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

class Solution {public:    int searchInsert(int A[], int n, int target)     {        if(n == 0)        return 0;      int low = 0;      int high = n - 1;      int mid;      while(low <= high)      {          mid = (low+high)/2;          if(A[mid] == target)            return mid;          else if(A[mid] < target)          {            low = mid+1;          }          else          {              high = mid - 1;          }      }      if(A[mid] > target)       return mid;      else       return mid + 1;          }};

8.Remove Duplicates from Sorted List

 Total Accepted: 25230 Total Submissions: 72036My Submissions

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

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/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {  public:      ListNode *deleteDuplicates(ListNode *head) {          // Start typing your C/C++ solution below          // DO NOT write int main() function          if (head == NULL)              return head;                    ListNode* cur = head;          ListNode* pre = head->next;          ListNode* del = NULL;          while(pre != NULL)          {              if (cur->val == pre->val)              {                  del = pre;                  pre = pre->next;                  cur->next = pre;                  delete del;              }              else              {                  cur = cur->next;                  pre = pre->next;              }          }          return head;      }  };