houserobber

compareVersionNumber/compareVersionNumber.cpp

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+// Source : https://oj.leetcode.com/problems/compare-version-numbers/
+
+/********************************************************************************** 
+ * 
+ * Compare two version numbers version1 and version1.
+ * If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
+ * 
+ * You may assume that the version strings are non-empty and contain only digits and the . character.
+ * The . character does not represent a decimal point and is used to separate number sequences.
+ * For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.
+ * 
+ * Here is an example of version numbers ordering:
+ * 0.1 < 1.1 < 1.2 < 13.37
+ * 
+ * Credits:Special thanks to @ts for adding this problem and creating all test cases.
+ *               
+ **********************************************************************************/
+
+class Solution {
+    public:
+        vector<int> &split(const string &s, char delim, vector<int> &elems) {
+            stringstream ss(s);
+            string item;
+            while (getline(ss, item, delim)) {
+                elems.push_back(atoi(item.c_str()));
+            }
+            return elems;
+        }
+
+
+        vector<int> split(const string &s, char delim) {
+            vector<int> elems;
+            split(s, delim, elems);
+            return elems;
+        }
+
+        void rightTrimZero(vector<int> &v){
+            while (v.back() == 0){
+                v.pop_back();
+            }
+        }
+
+        int compareVersion(string version1, string version2) {
+
+            //split the version by delimer '.'
+            vector<int> ver1 = split(version1, '.');
+            vector<int> ver2 = split(version2, '.');
+
+            //remove the right Zeros
+            rightTrimZero(ver1);
+            rightTrimZero(ver2);
+
+            //compare two versions 
+            for (int i=0; i<ver1.size() && i < ver2.size(); i++) {
+                if (ver1[i] < ver2[i]){
+                    return -1;
+                }else if(ver1[i] > ver2[i]){
+                    return 1;
+                }
+            }
+
+            //if the above for-loop is not returned, which means they are equal so far
+            //then check the length.
+            if (ver1.size() > ver2.size()) {
+                return 1;
+            }else if (ver1.size() < ver2.size()) {
+                return -1;
+            }
+
+            return 0;
+
+        }
+
+};

copyListWithRandomPointer/copyListWithRandomPointer.cpp

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+// Source : https://oj.leetcode.com/problems/copy-list-with-random-pointer/
+
+/********************************************************************************** 
+* 
+* A linked list is given such that each node contains an additional random pointer 
+* which could point to any node in the list or null.
+* 
+* Return a deep copy of the list.
+* 
+*               
+**********************************************************************************/
+
+
+
+/**
+ * Definition for singly-linked list with a random pointer.
+ * struct RandomListNode {
+ *     int label;
+ *     RandomListNode *next, *random;
+ *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
+ * };
+ */
+
+
+/*    
+ *
+ *  The idea as below:
+ *
+ *  Consider we have a linked list as below:
+ *
+ *    node1->random = node2;
+ *    node2->random = node1;
+ *    node3->random = node1;
+ *    
+ *       +-------------+                         
+ *       |             v                         
+ *    +-------+    +-------+    +-------+        
+ *    | node1 |----> node2 |----> node3 |--->NULL
+ *    +-------+    +-------+    +-------+        
+ *      ^  ^           |           |             
+ *      |  +-----------+           |             
+ *      +--------------------------+             
+ *    
+ *
+ *  To copy the list, 
+ *   
+ *    1) We insert a new node for each node's next position
+ *
+ * 
+ *       +-------------------------+                                     
+ *       |                         v                                     
+ *    +--+----+     +-----+    +-------+     +-----+    +-------+     +-----+     
+ *    | node1 |---> | NEW |----> node2 |---> | NEW |----> node3 |---> | NEW | ---> NULL
+ *    +-------+     +-----+    +---+---+     +-----+    +--+----+     +-----+ 
+ *      ^  ^                       |                       |             
+ *      |  +-----------------------+                       |             
+ *      +--------------------------------------------------+             
+ *
+ *    2) Then, we can construt the new node's random pointer:
+ *
+ *        (node1->next) -> random  = (node1->random) -> next;
+ *
+ *    3) After we take out all of the "NEW" node to finish the copy.
+ *    
+ */
+
+class Solution {
+public:
+    RandomListNode *copyRandomList(RandomListNode *head) {
+        RandomListNode *p = NULL, *h=NULL, *t=NULL;
+        if (head == NULL){
+            return NULL;
+        }
+
+        //creat a new node for each node and insert its next
+        p = head;
+        while ( p != NULL){
+            RandomListNode *node = new RandomListNode(p->label);
+            node->next = p->next;
+            p->next = node;
+            p = node->next;
+        }
+
+        //copy random pointer for each new node
+        p = head;
+        while (p != NULL){
+            if (p->random != NULL){
+                p->next->random = p->random->next;
+            }
+            p = p->next->next;
+        }
+
+        //break to two list
+        p = head;
+        h = t = head->next;
+        while ( p != NULL ){
+            p->next = p->next->next;
+            if (t->next!=NULL){
+                t->next = t->next->next;
+            }
+
+            p = p->next;
+            t = t->next;
+        }
+
+        return h;
+    }
+};
+
+
+/*
+ *  Considering we have a link as below:
+ *
+ *
+ *       +-------------+
+ *       |             v
+ *    +-------+    +-------+    +-------+
+ *    | node1 |----> node2 |----> node3 |--->NULL
+ *    +-------+    +-------+    +-------+
+ *      ^  ^           |           |
+ *      |  +-----------+           |
+ *      +--------------------------+
+ * 
+ *  1) Using a map to store each node's random pointer's position
+ *
+ *      map[node1->random] = 1;
+ *      map[node2->random] = 0;
+ *      map[node3->random] = 0;
+ *
+ *  2) Clone the linked list (only consider the next pointer)
+ *
+ *      new1 --> new2 --> new3 --> NULL 
+ *
+ *  3) Using an array to strore the order of the cloned linked-list
+ *
+ *      v[0] = &new1
+ *      v[1] = &new2
+ *      v[2] = &new3
+ * 
+ *  4) Then we can clone the random pointers.
+ *
+ *      new->random = v [ map[node->random] ]
+ *
+ */ 
+class MySolution {
+public:
+    RandomListNode *copyRandomList(RandomListNode *head) {
+
+        RandomListNode *p = NULL, *h=NULL, *t=NULL;
+        //using a map to store the random pointer's postion.
+        map<RandomListNode*, int> m;
+        //construct the map
+        int pos =0;
+        for ( p = head; p != NULL; p = p->next, pos++){
+            m[p] = pos;
+        }
+
+        //clone the linked list  (only consider the next pointer)
+        //and using a vector to store each node's postion.
+        vector<RandomListNode*> v;
+        for (p = head; p != NULL; p = p->next){
+            RandomListNode *node = new RandomListNode(p->label);
+            v.push_back(node);
+            if (h==NULL){
+                h = t = node;
+            }else{
+                t->next = node;
+                t = node;
+            }
+        }
+
+        //p => source link head 
+        //t => new link head
+        //move the p and t synchronously, and 
+        //     t->random = vector[ map[p->random] ];
+        for (t=h, p = head; t!=NULL && p!= NULL; p=p->next, t=t->next){
+            if (p->random!=NULL) {
+                pos = m[p->random];
+                t->random = v[pos];
+            }
+        }
+
+        return h;
+
+    }
+};