142. Linked List Cycle II

1. using map, but this will use extra space:(26ms)

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        unordered_map<ListNode*, int> m;
        ListNode* p=head;
        int cnt=0;
        while(p!=NULL){
            if(++m[p] == 2)return p;
            p=p->next;
        }
        return NULL;
    }
};



2. find out there is a circle first;
    then swipe each node to check whether there are in circle, the first in circle is the head of the circle.



3. using the 'find intersection in a merging lists' in http://xiamenhai.blogspot.com/2015/10/leetcode-intersection-of-two-linked.html.

Leetcode: Pascal's Triangle II

Given an index k, return the kth row of the Pascal's triangle.

For example, given k = 3,
Return [1,3,3,1].

Note:
Could you optimize your algorithm to use only O(k) extra space?

----------------------------

Note: O(k) means we can only use in-place way:

method: from right to left

eg.:

1   4   6  4  1 0 previous line

                            

0   0  0  0   0  1 now calculate in place from right to left 

0   0  0  0   5  1

0   0  0  10  5  1

0   0  10  10  5  1

0   5  10  10  5  1

1   5  10  10  5  1

Leetcode: Implement Stack using Queues

Implement Stack using Queues

My Submissions

Total Accepted: 28206 Total Submissions: 92981 Difficulty: Easy

Implement the following operations of a stack using queues.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• empty() -- Return whether the stack is empty.

Notes:

• You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid.
• Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
• You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

Update (2015-06-11):
The class name of the Java function had been updated to MyStack instead of Stack.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and all test cases.

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 Stack Design

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------------------------------------------------------------------------

stack.pop() using queue.pop():

using queue to push its front values to back until the original back one! Then pop the current front one.


////////////////////////////////////////////////////////////////////////////
class Stack {
public:
    queue<int> que;
    // Push element x onto stack.
    void push(int x) {
        que.push(x);
    }

    // Removes the element on top of the stack.
    void pop() {
        //using queue itself to push all front values to back until the last one
        for (int i=0;i<que.size()-1;i++){
            que.push(que.front());
            que.pop();
        }
        que.pop();
    }

    // Get the top element.
    int top() {
        return que.back();
    }

    // Return whether the stack is empty.
    bool empty() {
        return que.empty();
    }
};

Leetcode: Evaluate Reverse Polish Notation

Evaluate Reverse Polish Notation

My Submissions

Total Accepted: 57639 Total Submissions: 254722 Difficulty: Medium

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

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----------------------------------------------------------------------------------

思路：

典型的stack问题。逐一扫描每个token，如果是数字，则push入stack，如果是运算符，则从stack中pop出两个数字，进行运算，将结果push回stack。最后留在stack里的数即为最终结果。

以题中例子说明

exp:     2    1      +    3      *

stack:   2    2,1   3    3,3   9

///////////////////////////////////////////////////////////////////////

class Solution {

 public:

        int evalRPN(vector<string>& tokens) {

               stack<int> tokTmp;

               for (int i = 0; i<tokens.size(); i++){

                      //tokens[i] is a string, so "" is used!

                      if ((tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/")){

                            int y = tokTmp.top();

                            tokTmp.pop();

                            int x = tokTmp.top();

                            tokTmp.pop();

                            if (tokens[i] == "+")tokTmp.push(x + y);

                            else if (tokens[i] == "-")tokTmp.push(x - y);

                            else if (tokens[i] == "*")tokTmp.push(x*y);

                            else if (tokens[i] == "/")tokTmp.push(x / y);

                      }

                      else {

                            int x = stoi(tokens[i], nullptr, 10);

                            tokTmp.push(x);

                      }

               }

               return tokTmp.top();

        }

 };

Leetcode: Min Stack

 Min Stack

My Submissions

Total Accepted: 57566 Total Submissions: 271279 Difficulty: Easy

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

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--------------------------------------------------------------------------
Keep two stacks inside: one store all and the other store the minimal.
Note: only the current x is smaller or equal to the mimStack.top() is saved.



///////////////////////////////////////////////////////////
class MinStack {

public:
    void push(int x) {
        curSta.push(x);
        if(minSta.empty() || x<=minSta.top() )
           minSta.push(x);
    }

    void pop() {
       
        if(minSta.top() == curSta.top() )
            minSta.pop();
        curSta.pop(); //Note: this line should be put in last, otherwise, the above codes can not be compared!!!!       
    }

    int top() {
        return curSta.top();
    }

    int getMin() {
        return minSta.top();
    }
   
private:
    stack<int> minSta;
    stack<int> curSta;  
};

Leetcode: Permutations II

Permutations II

My Submissions

Total Accepted: 58417 Total Submissions: 216886 Difficulty: Medium

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

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 Backtracking

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-------------------------------------------------------------------------------------

BT.
Same as Permutations I. 
Note: 

vector<int>::iterator it = vt.begin();
vt.insert(it, nums[i]);
//insert a val before 'it'! In the end, it=vt.end(), (it)[0] will be wrong!!

SO use following:
vector<int>::iterator pre = it - 1;
vt.insert(pre, nums[i]);

////////////////////////////////////////////////////////////////////////

class Solution {

 public:

        void backTrac(vector<int>& nums, vector<vector<int>> &result, vector<vector<int>> &tmp, int i){

               if (i <= 0){ //when nums.size()==1

                      vector<vector<int>> tmp1;

                      vector<int> vTmp;

                      vTmp.push_back(nums[i]);

                      tmp1.push_back(vTmp);

                      tmp = tmp1;

                      result = tmp;//when nums.size()==1

                      return;

               }

               else{

                      backTrac(nums, result, tmp, i - 1);

                      //insertion

                      vector<vector<int>> tmp1;

                      for (int m = 0; m<tmp.size(); m++){

                            for (int n = 0; n <= tmp[m].size(); n++){

                                   vector<int> vt;

                                   vt = tmp[m];

                                   vector<int>::iterator it = vt.begin();

                                   it = it + n;

                                   if (i == nums.size()) break;

                                   //vt.insert(it, nums[i]);

                                   if (n == 0) vt.insert(it, nums[i]);

                                   else {

                                          vector<int>::iterator pre = it - 1;

                                          if (((pre)[0]) != nums[i])vt.insert(it, nums[i]); //insert a val before it! In the end, it=vt.end(), (it)[0] will be wrong!!

                                          else break;

                                   }

                                   tmp1.push_back(vt);

                            }

                      }

                      tmp = tmp1;

                      result = tmp;

               }

        }

        vector<vector<int>> permuteUnique(vector<int>& nums) {

               vector<vector<int>> result, tmp;

               if (nums.size() == 0)return result;

               int i = nums.size() - 1;

               backTrac(nums, result, tmp, i);

               return result;

        }

 };

Leetcode: Permutations

46. Permutations

My Submissions

Total Accepted: 81411 Total Submissions: 239626 Difficulty: Medium

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

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 Backtracking

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BT problem:
Using the similar method as SUBSET. Composing the subsets by insertion.




//////////////////////////////////////////////

class Solution {

 public:

        void backTrac(vector<int>& nums, vector<vector<int>> &result, vector<vector<int>> &tmp, int i){

               if (i == 0){

                      vector<vector<int>> tmp1;

                      vector<int> vTmp;

                      vTmp.push_back(nums[i]);

                      tmp1.push_back(vTmp);

                      tmp = tmp1;

                      result = tmp;

                      return;

               }

               else{

                      backTrac(nums, result, tmp, i - 1);

                      

                      //insertion

                      vector<vector<int>> tmp1;

                      for (int m = 0; m<tmp.size(); m++){

                            for (int n = 0; n<=tmp[m].size(); n++){

                                   vector<int> vt;

                                   vt = tmp[m];

                                   vector<int>::iterator it = vt.begin();

                                   it = it + n;

                                   if (i == nums.size()) break;

                                   vt.insert(it, nums[i]);

                                   tmp1.push_back(vt);

                            }

                      }

                      tmp = tmp1;

                      result = tmp;

               }

        }

        vector<vector<int>> permute(vector<int>& nums) {

               vector<vector<int>> result, tmp;

               int i = nums.size() - 1;

               backTrac(nums, result, tmp, i);

               return result;

        }

 };