LeetCode(86) -- 25, 73, 77

P25. Reverse Nodes in k-Group (Hard)

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed.

You may not alter the values in the list’s nodes, only nodes itself may be changed.

My Solution

public class P25_Reverse_Nodes_in_k_Group {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null) {
            return null;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode tail = findTail(head, k);
        ListNode before = dummy;
        while (tail != null) {
            ListNode after = tail.next;
            ListNode prev = before;
            ListNode current = head;
            ListNode next = current.next;
            while (current != after) {
                current.next = prev;
                prev = current;
                current = next;
                if (next != null) {
                    next = next.next;
                }
            }
            before.next = prev;
            before = head;
            head.next = after;
            head = head.next;
            tail = findTail(head, k);
        }
        return dummy.next;
    }

    private ListNode findTail(ListNode head, int k) {
        if (head == null) {
            return null;
        }
        int count = 1;
        ListNode current = head;
        while (current.next != null && count != k) {
            current = current.next;
            count++;
        }
        if (count == k) {
            return current;
        }
        return null;
    }
}

100%

The Best Solution

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode dum = dummy;
        int i = 0;
        while (head != null) {
            i++;
            if (i % k == 0) {
                dum = reverse(dum, head.next);
                head = dum.next;
            } else {
                head = head.next;
            }
        }
        return dummy.next;
    }
    
    public ListNode reverse(ListNode lastTail, ListNode end) {
        ListNode leadingHead = lastTail.next;
        ListNode returnNode = leadingHead;
        ListNode pre = lastTail;
        while (leadingHead != end) { // just normal reversing, like "Reverse Linked list I"
            ListNode then = leadingHead.next;
            
            leadingHead.next = pre;
            
            pre = leadingHead;
            leadingHead = then;
        }
        lastTail.next = pre;
        returnNode.next = end;
        return returnNode;
    }
}

P73. Set Matrix Zeroes (Medium)

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

My Solution

class Solution {
    public void setZeroes(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        boolean[] row = new boolean[m];
        boolean[] col = new boolean[n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0) {
                    row[i] = true;
                    col[j] = true;
                }
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (row[i] || col[j]) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
}

100%

P77. Combinations (Medium)

Given two integers n and k, return all possible combinations of k numbers out of 1 … n.

My Solution

class Solution {
    List<List<Integer>> ans = new ArrayList<>();
    public List<List<Integer>> combine(int n, int k) {
        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = i + 1;
        }
        backtrack(nums, 0, new ArrayList<>(), k);
        return ans;
    }

    private void backtrack(int[] nums, int start, List<Integer> tempList, int k) {
        if (tempList.size() == k) {
            ans.add(new ArrayList<>(tempList));
            return;
        }
        for (int i = start; i < nums.length; i++) {
            tempList.add(nums[i]);
            backtrack(nums, i + 1, tempList, k);
            tempList.remove(tempList.size() - 1);
        }
    }
}

70%

The Best Solution

class Solution {
    List<List<Integer>> list = new ArrayList<>();
    public List<List<Integer>> combine(int n, int k){
        int[] array = new int[n];
        for(int i=1; i<=n; i++){
            array[i-1] = i;
        }
        List<Integer> l = new ArrayList<>();
        recursiveCombine(array, 0, k, l);
        return list;
    }

    private void recursiveCombine(int[] array, int currCursor, int leftK, List<Integer> l) {
        if(leftK==0){
            list.add(new ArrayList<>(l));
            return;
        }
        for(int i=currCursor; i<=array.length-leftK; i++){
            l.add(array[i]);
            recursiveCombine(array, i+1, leftK -1, l);
            l.remove(l.size()-1);
        }
    }
}

进得越深，需要做的loop次数越少。