Insert in Cyclic Linked List

/***
Given a node from a cyclic linked list which has been sorted, write a function to insert a value into the
list such that it remains a cyclic sorted list. The given node can be any single node in the list.
Pay attention on edge case.
1. input is nullptr
2. duplicate and single node
http://leetcode.com/2011/08/insert-into-a-cyclic-sorted-list.html
***/
#include <iostream>
using namespace std;

struct node {
       int val;
       node *next;
       node(int v) {
              this->val = v;
              this->next = nullptr;
       }
};


void insert(node *&aNode, int x) {
       if(aNode == nullptr) {           
              aNode = new node(x);
              aNode->next = aNode;
              return;
       }

       node *n = aNode;
       while(true) {
              if(n->val > n->next->val) { // n is tail and n->next is head
                     if(n->val < x || n->next->val > x) break;
                     else n = n->next;
              } else if(n->val < n->next->val) { // n is mid node
                     if(n->val < x && n->next->val > x) break;
                     else n = n->next;   
              } else { // n is duplicate or itself
                     if(n == aNode) break;
                     else n = n->next;
              }
       }
       node *k = new node(x);
       node *t = n->next;
       n->next = k;
       k->next = t;
}

void main() {
       node *head = new node(0);
//     node *head = nullptr;
       node *p, *a;
       p = head;
//     a = p;
       for (int i = 0; i<1; i += 2)
       {
              //if (i == 6) a = p;
              node *n = new node(3);
              p->next = n;
              p = p->next;
       }

       head = head->next;
       p->next = head;

       insert(a, 5);

       p = head;
       do {
              cout<<p->val;
              p = p->next;
       } while (p != head);

}



Comments

Post a Comment

Popular posts from this blog

Sudoku

/*** Sudoku Given a partially filled 9×9 2D array ‘grid[9][9]‘, the goal is to assign digits (from 1 to 9) to the empty cells so that every row,   column, and subgrid of size 3×3 contains exactly one instance of the digits from 1 to 9. Backtracking Algorithm Like all other Backtracking problems, we can solve Sudoku by one by one assigning numbers to empty cells.  Before assigning a number, we check whether it is safe to assign. We basically check that the same number  is not present in current row, current column and current 3X3 subgrid. After checking for safety,  we assign the number, and recursively check whether this assignment leads to a solution or not.  If the assignment doesn’t lead to a solution, then we try next number for current empty cell.  And if none of number (1 to 9) lead to solution, we return false. ***/ #include <iostream> using namespace std; class Sudoku { public :      ...
Read more

Longest prefix matching - trie

/*** Longest prefix matching Given a dictionary of words and an input string, find the longest prefix of the string which is also a word in dictionary. Example: Let the dictionary contains the following words: {are, area, base, cat, cater, children, basement} Below are some input/output examples: -------------------------------------- Input String            Output -------------------------------------- caterer                 cater basemexy                base child                   < Empty > ***/ #include <iostream> #include <string> using namespace std; struct node {        char val; ...
Read more

Climbing Stairs

You are climbing a stair case. It takes  n  steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note:  Given  n  will be a positive integer. class Solution { public:     int climbStairs(int n) {         if (n == 0) return 0;                  if (n == 1) return 1;                  // dp space complexity is O(n)         vector<int> dp(n);         dp[0] = 1;         dp[1] = 2;                  for(int i = 2; i<n; i++)         {             dp[i] = dp[i-1] + dp[i-2];         }                  return dp[ n-1 ]...
Read more