Leecode 收费题

Q. 156: Binary Tree Upside Down 

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:
Given a binary tree {1,2,3,4,5},

    1   / \  2   3 / \4   5

return the root of the binary tree [4,5,2,#,#,3,1].

   4  / \ 5   2    / \   3   1  

C++实现如下:

<span style="color:#333333;">TreeNode *upsideDownBinaryTree(TreeNode *root) {  TreeNode *new_root = NULL;  buildUpsideDownBT(root, new_root);  return new_root;}TreeNode *buildTree(TreeNode *root, TreeNode* &new_root) {  if(!root) return root;  if(!root->left && !root->right) {    new_root = root;    return root;  }  TreeNode *parent = buildTree(root->left, new_root);  parent->left = root->right;  parent->right = root;  root->left = root->right = NULL;  //very important  return parent->right;}

Q. 157: Read N Characters Given Read4 

The API: int read4(char *buf) reads 4 characters at a time from a file.

The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.

By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file.

C++实现如下:

int read4(char* buff);int read(char* buff, int n){  int total_len = 0;  int curr_len = 0;  while(total_len+4<=n){    curr_len = read4(buff+total_len);    total_len += curr_len;    if(curr_len<4) break;  }  if(curr_len<4 && total_len==n)  return total_len;  char* new_buff = new char[5];  curr_len = read4(new_buff);  curr_len = min(curr_len, n-total_len);  for(int i=0; i<curr_len; ++i) buff[total_len+i] = new_buff[i];    delete new_buff;    return total_len+curr_len;}

Q. 158: Read N Characters Given Read4 II - Call multiple times

The API: int read4(char *buf) reads 4 characters at a time from a file.

The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.

By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file.

分析: 注意这个可以读多次,如果buffer有剩余,下次要从剩下的地方读取.

C++实现如下:

int read(char* buff, int n){  int total_len = 0;  static int curr_len = 4;  static char* local_buff = new char[5];  static int local_p = 4;  while(total_len<n){    if(local_p==curr_len)      { if(curr_len<4) break;curr_len = read4(local_buff);local_p = 0;      }    if(local_p < curr_len) buff[total_len++] = local_buff[local_p++];  }  return total_len;}

Q. 159: Longest Substring with At Most Two Distinct Characters 

Given a string, find the length of the longest substring T that contains at most 2 distinct characters.

For example, Given s = “eceba”, T is "ece" which its length is 3.

分析: 其实更一般的情况是对任意n个distinct characters求最大字串.

C++实现如下:

int longestSubstring(const string& s, const int n){  int hash[128];  memset(hash, 0, sizeof(int)*128);  int left = 0;  int right = 0;  int k = 0; //number of distinct character  int max_len = 0;   while(right<s.size()){    ++hash[s[right]];    if(hash[s[right]]==1) ++k;    ++right;        while(k>n){      --hash[s[left]];      if(hash[s[left]]==0) --k;      ++left;    }    if(max_len<right-left) max_len = right - left;    cout << left << ", " << right << ", " << k << endl;  }    return max_len;}

Q. 161: One Edit Distance  

Given two strings S and T, determine if they are both one edit distance apart.

Hint:
1. If | n – m | is greater than 1, we know immediately both are not one-edit distance apart.
2. It might help if you consider these cases separately, m == n and m ≠ n.
3. Assume that m is always ≤ n, which greatly simplifies the conditional statements. If m > n, we could just simply swap S and T.
4. If m == n, it becomes finding if there is exactly one modified operation. If m ≠ n, you do not have to consider the delete operation. Just consider the insert operation in T.

分析:  For one edit distance, there are two qualified cases. Case 1 is that S.length equals T.length, so we just find out that if there is only one different character between S and T. Case 2 is that T is one character longer than S, except that different character, all other characters are the same.

Java实现如下:

public boolean isOneEditDistance(String s, String t) {   int lenS = s.length(), lenT = t.length();      //if s is longer than t, switch them.   //then we can make sure that t is always   //longer than s.   if (lenS > lenT)       return isOneEditDistance(t, s);      //if the length difference is larger than 1,   //obviously the edit distance cannot be 1   int lengthDiff = lenT - lenS;   if (lengthDiff > 1)       return false;      //iterate though shorter string s, find the first   //different char between s and t.   int i = 0;   while (i < lenS && s.charAt(i) == t.charAt(i))       i++;      //when i equals lenS, there are two possibilities   //one is that s equals t(lengthDiff = 0), return false;   //otherwise it is the one edit distance, return true.   if (i == lenS)       return lengthDiff > 0;      //case: bcd, ccd       //case 1 that i stops in the middle or beginning of s,   //skip this different character.       if (lengthDiff == 0)       i++;      //skip the different character in longer string t. case: bcd, ccde   while (i < lenS && s.charAt(i) == t.charAt(i + lengthDiff))       i++;      return i == lenS;}

Q.163: Missing Ranges

Given a sorted integer array where the range of elements are [lower, upper] inclusive, return its missing ranges.

For example, given [0, 1, 3, 50, 75], lower = 0 and upper = 99, return ["2", "4->49", "51->74", "76->99"].

C++实现如下:

vector<string> missingRanges(const vector<int>& nums, int start, int end){  int missing = start;  vector<string> result;  for(int i=0; i<nums.size(); ++i){    if(missing<nums[i]){string s = to_string(missing);if(missing<nums[i]-1){  s += "->";  s += to_string(nums[i]-1);}result.push_back(s);missing = nums[i]+1;    }    else ++missing;  }      if(missing<=end){    string s = to_string(missing);    if(missing<end){      s += "->";      s += to_string(end);    }    result.push_back(s);      }  return result;}

Q.167: Two Sum II - Input array is sorted 

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

C++实现如下:

pair<int, int> twoSum(const vector<int>& nums, int target){      int left = 0;  int right = nums.size()-1;  while(left<right){    if(nums[left]+nums[right]<target) ++left;    else if(nums[left]+nums[right]>target) --right;    else break;  }  return make_pair(left, right);}

Q. 170: Two Sum III - Data structure design 

Design and implement a TwoSum class. It should support the following operations: add and find.

add - Add the number to an internal data structure.
find - Find if there exists any pair of numbers which sum is equal to the value.

For example,

add(1); add(3); add(5);find(4) -> truefind(7) -> false

C++实现如下:

class TwoSum{  unordered_map<int, int> hash;public:  void add(int number){    ++hash[number];  }  bool find(int target){    for(unordered_map<int, int>::iterator it=hash.begin(); it!=hash.end(); ++it){      if(hash.count(target - it->first)){if(target != it->first*2) return true;else if(hash[it->first]>1) return true;      }    }    return false;  }  };

Q. 186: Reverse Words in a String II 

 Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.
The input string does not contain leading or trailing spaces and the words are always separated by a single space.
For example,
Given s = "the sky is blue",
return "blue is sky the".
Could you do it in-place without allocating extra space?

C++实现如下:

string reverseWords(string s){  reverse(s.begin(), s.end());  size_t start = 0;  size_t end = s.find(' ');  while(end!=string::npos){        reverse(s.begin()+start, s.begin()+end);    start = end+1;    end = s.find(' ', start);  }    reverse(s.begin()+start, s.end());  return s;}