hash & heap - 20181023 - 20181026

129. Rehashing

 /**
  * Definition for ListNode
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */

 public class Solution {
     /**
      * @param hashTable: A list of The first node of linked list
      * @return: A list of The first node of linked list which have twice size
      */
     public ListNode[] rehashing(ListNode[] hashTable) {
         // write your code here
         if (hashTable.length == 0) {
             return hashTable;
         }
         int capacity = hashTable.length * 2;
         ListNode[] result = new ListNode[capacity];
         for (ListNode listNode : hashTable) {
             while (listNode != null) {
                 int newIndex = (listNode.val % capacity + capacity) % capacity;
                 if (result[newIndex] == null) {
                     result[newIndex] = new ListNode(listNode.val);
                 } else {
                     ListNode head = result[newIndex];
                     while (head.next != null) {
                         head = head.next;
                     }
                     head.next = new ListNode(listNode.val);
                 }
                 listNode = listNode.next;
             }
         }
         return result;
     }
 };

134. LRU Cache

 class Node {
     int key;
     int value;
     Node next;

     Node(int key, int value) {
         this.key = key;
         this.value = value;
         this.next = null;
     }
 }

 public class LRUCache {
     int size;
     int capacity;
     Node tail;
     Node dummy;
     Map<Integer, Node> keyToPrevMap;

     /*
     * @param capacity: An integer
     */
     public LRUCache(int capacity) {
         // do intialization if necessary
         this.capacity = capacity;
         this.size = 0;
         this.dummy = new Node(0, 0);
         this.tail = dummy;
         keyToPrevMap = new HashMap<>();
     }

     public void moveRecentNodeToTail(int key) {
         Node prev = keyToPrevMap.get(key);
         Node curr = prev.next;
         if (curr == tail) {
             return;
         }

         prev.next = curr.next;
         tail.next = curr;
         keyToPrevMap.put(key, tail);
         //删掉现节点之后，不仅要改变前指针指向，还需要更新map
         if(prev.next!=null){
             keyToPrevMap.put(prev.next.key,prev);
         }
         tail = curr;
     }

     /*
      * @param key: An integer
      * @return: An integer
      */
     public int get(int key) {
         // write your code here
         if (keyToPrevMap.containsKey(key)) {
             moveRecentNodeToTail(key);
             return tail.value;
         }
         return -1;
     }

     /*
      * @param key: An integer
      * @param value: An integer
      * @return: nothing
      */
     public void set(int key, int value) {
         // write your code here
         if (get(key) != -1) {
             Node prev = keyToPrevMap.get(key);
             prev.next.value = value;
             return;
         }

         //无此值 set并放在队尾
         if (size < capacity) {
             size++;
             Node newNode = new Node(key, value);
             tail.next = newNode;
             keyToPrevMap.put(key, tail);
             tail = newNode;
             return;
         }

         //空间已满，需要删掉头指针: 可以直接替换头指针key-value值，再移到队尾来实现
         Node first = dummy.next;
         keyToPrevMap.remove(first.key);
         first.key = key;
         first.value = value;
         keyToPrevMap.put(key, dummy);
         moveRecentNodeToTail(key);
     }
 }

4. Ugly Number II

method1：用优先队列

 public class Solution {
     /**
      * @param n: An integer
      * @return: the nth prime number as description.
      */
     public int nthUglyNumber(int n) {
         // write your code here
         Queue<Long> queue = new PriorityQueue<>();
         Set<Long> set = new HashSet<>();
         int[] factors = new int[3];

         factors[0] = 2;
         factors[1] = 3;
         factors[2] = 5;
         queue.add(1L);
         Long number = 0L;
         for (int i = 0; i < n; i++) {
             number = queue.poll();
             for (int j = 0; j < 3; j++) {
                 if (!set.contains(number * factors[j])) {
                     queue.add(number * factors[j]);
                     set.add(number * factors[j]);
                 }
             }
         }
         return number.intValue();
     }
 }

method2: 丑数一定是某一个丑数*2或者*3或者*5的结果，维护三个指针，例如如果该指针*2结果大于上一个丑数，指针+1

 public class Solution {
     /**
      * @param n: An integer
      * @return: the nth prime number as description.
      */
     public int nthUglyNumber(int n) {
         // write your code here
         List<Integer> res = new ArrayList<>();
         res.add(1);
         int p2 = 0;
         int p3 = 0;
         int p5 = 0;
         for (int i = 1; i < n; i++) {
             int lastNum = res.get(res.size() - 1);
             while (res.get(p2) * 2 <= lastNum) p2++;
             while (res.get(p3) * 3 <= lastNum) p3++;
             while (res.get(p5) * 5 <= lastNum) p5++;
             res.add(Math.min(Math.min(res.get(p2) * 2, res.get(p3) * 3), res.get(p5) * 5));
         }

         return res.get(n - 1);
     }
 }

545. Top k Largest Numbers II

 public class Solution {
     private int maxSize;
     private Queue<Integer> minHeap;

     /*
     * @param k: An integer
     */
     public Solution(int k) {
         // do intialization if necessary
         this.maxSize = k;
         this.minHeap = new PriorityQueue<>();
     }

     /*
      * @param num: Number to be added
      * @return: nothing
      */
     public void add(int num) {
         // write your code here
         if (minHeap.size() < maxSize) {
             minHeap.offer(num);
             return;
         }
         if (num > minHeap.peek()) {
             minHeap.poll();
             minHeap.offer(num);
         }
     }

     /*
      * @return: Top k element
      */
     public List<Integer> topk() {
         // write your code here
         List<Integer> res = new ArrayList<>();
         for (Integer num : minHeap) {
             res.add(num);
         }
         Collections.sort(res);
         Collections.reverse(res);
         return res;
     }
 }

104. Merge K Sorted Lists

method1:利用最小堆

 /**
  * Definition for ListNode.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int val) {
  *         this.val = val;
  *         this.next = null;
  *     }
  * }
  */
 public class Solution {

     private Comparator<ListNode> listNodeComparator = new Comparator<ListNode>() {
         @Override
         public int compare(ListNode o1, ListNode o2) {
             return (o1.val - o2.val);
         }
     };

     /**
      * @param lists: a list of ListNode
      * @return: The head of one sorted list.
      */
     public ListNode mergeKLists(List<ListNode> lists) {
         // write your code here
         if (lists == null || lists.size() == 0) {
             return null;
         }

         Queue<ListNode> minheap = new PriorityQueue<>(lists.size(), listNodeComparator);
         for (ListNode node : lists) {
             if(node != null){
                 minheap.add(node);
             }
         }

         ListNode dummy = new ListNode(0);
         ListNode tail = dummy;
         while (!minheap.isEmpty()) {
             ListNode head = minheap.poll();
             tail.next = head;
             tail = head;
             if (head.next != null) {
                 minheap.add(head.next);
             }
         }
         return dummy.next;
     }
 }

method2: merge 2 by 2（待二刷）

method3: 分治（待二刷）

613. High Five

 /**
  * Definition for a Record
  * class Record {
  *     public int id, score;
  *     public Record(int id, int score){
  *         this.id = id;
  *         this.score = score;
  *     }
  * }
  */
 public class Solution {
     /**
      * @param results a list of <student_id, score>
      * @return find the average of 5 highest scores for each person
      * Map<Integer, Double> (student_id, average_score)
      */
     public Map<Integer, Double> highFive(Record[] results) {
         // Write your code here
         Map<Integer, Double> resMap = new HashMap<>();
         if (results == null || results.length == 0) {
             return resMap;
         }
         Map<Integer, PriorityQueue<Integer>> scoreMap = new HashMap<>();
         for (Record record : results) {
             if (!scoreMap.containsKey(record.id)) {
                 scoreMap.put(record.id, new PriorityQueue<Integer>());
             }

             PriorityQueue<Integer> heap = scoreMap.get(record.id);
             if (heap.size() < 5) {
                 heap.add(record.score);
                 continue;
             }

             if (record.score > heap.peek()) {
                 heap.poll();
                 heap.add(record.score);
             }
         }

         for (Map.Entry<Integer, PriorityQueue<Integer>> entry : scoreMap.entrySet()) {
             PriorityQueue<Integer> scoreHeap = entry.getValue();
             int sum = 0;
             while (!scoreHeap.isEmpty()) {
                 sum += scoreHeap.poll();
             }
             double ave = sum / 5.0;
             resMap.put(entry.getKey(), ave);
         }

         return resMap;

     }
 }

612. K Closest Points

 public class Solution {
     /**
      * @param points: a list of points
      * @param origin: a point
      * @param k:      An integer
      * @return: the k closest points
      */
     public Point[] kClosest(Point[] points, Point origin, int k) {
         // write your code here
         if (points == null || points.length < k || k == 0) {
             return null;
         }

         Comparator<Point> comparator = new Comparator<Point>() {
             @Override
             public int compare(Point o1, Point o2) {
                 return compareDistance(o2, o1, origin);
             }
         };
         Point[] res = new Point[k];
         PriorityQueue<Point> maxHeap = new PriorityQueue<>(comparator);
         for (Point point : points) {
             if (maxHeap.size() < k) {
                 maxHeap.add(point);
                 continue;
             }
             if (compareDistance(point, maxHeap.peek(), origin) < 0) {
                 maxHeap.poll();
                 maxHeap.add(point);
             }
         }

         for (int i = k - 1; i >= 0; i--) {
             res[i] = maxHeap.poll();
         }
         return res;
     }

     private int compareDistance(Point o1, Point o2, Point center) {
         int distance1 = (o1.x - center.x) * (o1.x - center.x) + (o1.y - center.y) * (o1.y - center.y);
         int distance2 = (o2.x - center.x) * (o2.x - center.x) + (o2.y - center.y) * (o2.y - center.y);
         if (distance1 == distance2) {
             return o1.x != o2.x ? o1.x - o2.x : o1.y - o2.y;
         }
         return distance1 - distance2;
     }
 }

81. Find Median from Data Stream

 public class Solution {
     /**
      * @param nums: A list of integers
      * @return: the median of numbers
      */

     //此题中位数定义和一般的有区别，写的时候需要注意
     //存储后半部分的数据
     private PriorityQueue<Integer> minHeap;
     //存储前半部分的数据
     private PriorityQueue<Integer> maxHeap;
     private int count;//已加入的数

     public int[] medianII(int[] nums) {
         // write your code here
         if (nums == null || nums.length == 0) {
             return null;
         }
         int len = nums.length;
         int[] res = new int[len];
         Comparator<Integer> maxComparator = new Comparator<Integer>() {
             @Override
             public int compare(Integer o1, Integer o2) {
                 return o2.compareTo(o1);
             }
         };
         minHeap = new PriorityQueue<>(len);
         maxHeap = new PriorityQueue<>(len, maxComparator);
         for (int i = 0; i < len; i++) {
             addNumber(nums[i]);
             res[i] = getMedian();
         }

         return res;
     }

     public void addNumber(int num) {
         maxHeap.add(num);
         count++;
         //总是先add前半部分数，保证maxHeap.size()>=minHeap.size()
         if (count % 2 == 1) {
             if (minHeap.isEmpty()) {
                 return;
             }

             //交换堆顶，保证有序
             if (maxHeap.peek() > minHeap.peek()) {
                 int maxPeek = maxHeap.poll();
                 int minPeek = minHeap.poll();
                 minHeap.add(maxPeek);
                 maxHeap.add(minPeek);
             }

             return;
         }
         //当count为奇数时，maxHeap.size-minHeap.size=1，因而当count变为偶数时，只用无脑再将peek取出添到后半部分
         minHeap.add(maxHeap.poll());
         return;
     }

     public int getMedian() {
         return maxHeap.peek();
     }
 }

401. Kth Smallest Number in Sorted Matrix

此题注意应该维护的是最小堆，很容易误判成最大堆

 class Pair {
     private int x;
     private int y;
     private int val;

     public Pair(int x, int y, int val) {
         this.x = x;
         this.y = y;
         this.val = val;
     }

     public int getX() {
         return x;
     }

     public void setX(int x) {
         this.x = x;
     }

     public int getY() {
         return y;
     }

     public void setY(int y) {
         this.y = y;
     }

     public int getVal() {
         return val;
     }

     public void setVal(int val) {
         this.val = val;
     }
 }

 public class Solution {
     /**
      * @param matrix: a matrix of integers
      * @param k:      An integer
      * @return: the kth smallest number in the matrix
      */
     public int kthSmallest(int[][] matrix, int k) {
         // write your code here
         if (matrix == null || matrix.length == 0 || matrix.length * matrix[0].length < k) {
             return -1;
         }
         Comparator<Pair> comparator = new Comparator<Pair>() {
             @Override
             public int compare(Pair o1, Pair o2) {
                 return o1.getVal() - o2.getVal();
             }
         };
         int r = matrix.length;
         int c = matrix[0].length;
         PriorityQueue<Pair> minHeap = new PriorityQueue<>(comparator);
         boolean[][] visited = new boolean[r][c];

         minHeap.add(new Pair(0, 0, matrix[0][0]));
         visited[0][0] = true;

         for (int i = 1; i <= k - 1; i++) {
             Pair cur = minHeap.poll();
             if (cur.getX() + 1 < r && !visited[cur.getX() + 1][cur.getY()]) {
                 minHeap.add(new Pair(cur.getX() + 1, cur.getY(), matrix[cur.getX() + 1][cur.getY()]));
                 visited[cur.getX() + 1][cur.getY()] = true;
             }
             if (cur.getY() + 1 < c && !visited[cur.getX()][cur.getY() + 1]) {
                 minHeap.add(new Pair(cur.getX(), cur.getY() + 1, matrix[cur.getX()][cur.getY() + 1]));
                 visited[cur.getX()][cur.getY() + 1] = true;
             }
         }

         return minHeap.peek().getVal();
     }
 }