13. Roman to Integer

Given a roman numeral, convert it to an integer.

 

Input is guaranteed to be within the range from 1 to 3999.

 

解题思路：

 

 

 

屏幕剪辑的捕获时间: 2016/3/4 22:52

从左到右，两两进行比较。。。使用map和数组标记两种方式所消耗的时间不同。。

#include<string> #include<iostream> #include<map> using namespace std; class Solution { public:     int romanToInt(string s) {         int n=s.size();         if(n==0)return 0;         // 68ms         //map<char,int>m;         //m['I']=1;         //m['X']=10;         //m['C']=100;         //m['M']=1000;         //m['V']=5;         //m['L']=50;         //m['D']=500;           // 32ms         int m[26]={0};         m['I'-'A']=1;         m['X'-'A']=10;         m['C'-'A']=100;         m['M'-'A']=1000;         m['V'-'A']=5;         m['L'-'A']=50;         m['D'-'A']=500;         int ans=0;         for(int i=0;i<n-1;i++){             int a=m[s[i]-'A'];             if(s[i]=='I'||s[i]=='X'||s[i]=='C'){             int b=m[s[i+1]-'A'];             if(a<b)ans-=a;             else                 ans+=a;             }             else                 ans+=a;         }         ans+=m[s[n-1]-'A'];         return ans;     } };

 

235. Lowest Common Ancestor of a Binary Search Tree

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

 

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

 

        _______6______

       /              \

    ___2__          ___8__

   /      \        /      \

   0      _4       7       9

         /  \

         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

 

解题思路：

因为是二叉搜索树，有左子树的所有值都小于父节点的值，右子树的所有值都大于父节点，因此递归判断两个节点在左子树还是右子树，若在两边，则父节点就是其最小公共祖先，否则递归搜索同在一遍的子树。

#include<windows.h> using namespace std; struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {} };    class Solution { public: // 40ms-42ms     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {         if(root==NULL)return root;         if(p==NULL)return q;         if(q==NULL)return p;         int val=root->val;         if((val-p->val)*(val-q->val)<=0)return root;         if(val>p->val)return lowestCommonAncestor(root->left,p,q);         return lowestCommonAncestor(root->right,p,q);     } };

 

191. Number of 1 Bits

Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

 

For example, the 32-bit integer '11' has binary representation 00000000000000000000000000001011, so the function should return 3.

 

解题思路：

1、bit manipulate  利用n&(n-1)快速去掉最低位的1.。

2、使用c++ stl中的bitset类型，将其转换为32位的二进制，利用count方法得到1的个数

3、已知输入为32位无符号的整数，可以知道只要循环32次，求每位是否为1.

#include<stdint.h> #include<bitset> using namespace std; class Solution { public:     // 8ms   //  int hammingWeight(uint32_t n) {         //bitset<32>ans(n);         //return ans.count();   //  }     // 8ms     int hammingWeight(uint32_t n) {         int cnt=0;         for(;n;n>>=1){             if(n&1)cnt++;         }         return cnt;     }     // 4ms     int hammingWeight(uint32_t n) {         int cnt = 0;         for(; n; n >>= 1)             if(n&1) ++cnt;         return cnt;     } };

 

144. Binary Tree Preorder Traversal

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].

 

解题思路：

题目很简单，就是求一个二叉树的前序遍历。。使用栈实现递归过程。。。这里分别给出保存TreeNode的栈和保存TreeNode指针的栈。。

/**  * 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:     vector<int> preorderTraversal(TreeNode* root) {         vector<int>res;         stack<TreeNode>s;         if(root==NULL)return res;         s.push(*(root));         while(!s.empty()){             TreeNode tmp=s.top();             s.pop();             res.push_back(tmp.val);             if(tmp.right!=NULL)s.push(*(tmp.right));             if(tmp.left!=NULL)s.push(*(tmp.left));         }         return res;     } };

 

/**  * 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:     vector<int> preorderTraversal(TreeNode* root) {         vector<int>res;         stack<TreeNode*>s;         if(root==NULL)return res;         s.push(root);         while(!s.empty()){             TreeNode* tmp=s.top();             s.pop();             res.push_back(tmp->val);             if(tmp->right!=NULL)s.push(tmp->right);             if(tmp->left!=NULL)s.push(tmp->left);         }         return res;     } };

 

java如果使用栈或者递归方法，耗时2ms，在discuss那里看到不用stack，用rights数组保存遍历过程中遇到的右结点，当当前结点的左结点为空，则从rights数组中找到最近的右结点。。。该方法耗费空间。。

/**  * Definition for binary tree  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode(int x) { val = x; }  * }  */ public class Solution {    public List<Integer> preorderTraversal(TreeNode root) {             List<Integer>order=new ArrayList<Integer>();             List<TreeNode>rights=new ArrayList<TreeNode>();             if(root==null)return order;             TreeNode cur=root;             int ind=-1;             while(cur!=null){                 order.add(cur.val);                 if(cur.right!=null){                     ind++;                     rights.add(ind,cur.right);                 }                 if(cur.left!=null)cur=cur.left;                 else{                     cur=ind<0?null:rights.get(ind);                     ind--;                 }                               }             return order;         } }

 

 

206. Reverse Linked List

Reverse a singly linked list.

 

解题思路：

求一个链表的逆序。一种方法是遍历链表，将节点保存到stack中，在一个个出栈加到链表中；另一个方法是遍历链表，将每个节点插入到队首。。

 

C++ 8ms

/**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     ListNode *next;  *     ListNode(int x) : val(x), next(NULL) {}  * };  */ class Solution { public:     ListNode* reverseList(ListNode* head) {         if(head==NULL)return head;         ListNode* root=new ListNode(0);         ListNode* p=head;         while(p!=NULL){             ListNode*q=p->next;             p->next=root->next;             root->next=p;             p=q;         }         return root->next;     } };

 

Java 0ms

/**  * Definition for singly-linked list.  * public class ListNode {  *     int val;  *     ListNode next;  *     ListNode(int x) { val = x; }  * }  */ public class Solution {    public ListNode reverseList(ListNode head) {         if(head==null)return head;         ListNode p=head;         ListNode root=new ListNode(0);         while(p!=null){             ListNode q=p.next;             p.next=root.next;             root.next=p;             p=q;         }         return root.next;     } }

 

3ms

public class Solution {    public ListNode reverseList(ListNode head) {         if(head==null)return head;         Stack<Integer>stack=new Stack<Integer>();         ListNode p=head;         while(p!=null){             stack.push(p.val);             p=p.next;         }         p=head;         while(p!=null){             p.val=stack.pop();             p=p.next;         }         return head;     } }