1. Reverse Nodes in k-Group

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.

思路：嘗試遍歷鏈表，根據索引關系來逆置指定部分，結果發現復雜無比，gg。還是看了下解析，置頂的方法是用遞歸來解題。其基本想法是對於找到的k個節點逆置，然後對剩下的節點作相同處理。雖然不是很高效但是卻非常通俗易懂。

代碼：

public ListNode reverseKGroup(ListNode head, int k){
    ListNode cur=head;
    int count=0;
    while(cur!=null && count<k){
        cur=cur.next;
        count++;
    }
    
    
 
if(count==k){
        cur=reverseKGroup(cur, k); //標記1
        //這裏的逆轉邏輯是依次將第一個移動到相應後面的位置，cur始終指向這個位置的下一個
        while(count-- > 0){
            ListNode temp=head.next; //將當前頭結點的下一個賦給temp
            head.next=cur; //cur指向下一組逆轉置後的頭結點(第一個幾點)，將head移動到逆置後的位置，也就是cur前面
            cur=head; //因為head已經移動好了，接下來移動temp的話，應該是移動到head的前面，所以需要將cur指向head
            head=temp；//將原先第二個節點temp置為新的head，循環往後移，直至移動了k個            
        }
        //逆置完成後cur應該指向當前k個節點中的第一個，
        //需要將其賦給head返回，標記1處的cur才能指向其後面k個節點逆置後的第一個節點
        head=cur;
    }
    return head;
}

2. Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

1. Each of the digits 1-9 must occur exactly once in each row.
2. Each of the digits 1-9 must occur exactly once in each column.
3. Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character ‘.‘.

A sudoku puzzle...

...and its solution numbers marked in red.

Note:

• The given board contain only digits 1-9 and the character ‘.‘.
• You may assume that the given Sudoku puzzle will have a single unique solution.
• The given board size is always 9x9.

思路：看了下discussion，發現有多種解題方法，比如DFS或者Backtracing，也有算法大神給出了很厲害的解法。還是選擇較為簡單的Backtracing方法來考慮。其基本思路是：

1. 遍歷整個數獨，對於空缺的部分，嘗試用1到9填充。

2. 填充前先判斷當前位置所對應的行和列，以及所在的3*3矩陣有沒有整個1~9數字，如果有則跳過判斷1~9中的下一個。

3. 填充好了後，將當前矩陣作為新的輸入重復步驟1，2，3，這樣遞歸，如果有一層如果1~9都填不進去則表示數獨不合法，上層填的有誤，需返回false到上層重置上層填的數字。如果順利遞歸到最底層，也就是填好了整個數獨，因為每次的賦值標準是數獨合法，所以填好的數獨也是合法的，故返回true。

代碼：

public class Solution {
    public void solveSudoku(char[][] board) {
        if(board == null || board.length == 0)
            return;
        solve(board);
    }
    
    public boolean solve(char[][] board){
        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(board[i][j] == ‘.‘){
                    for(char c = ‘1‘; c <= ‘9‘; c++){//trial. Try 1 through 9,註意這裏直接c++就可以取到字符數字的下一個
                        if(isValid(board, i, j, c)){
                            board[i][j] = c; //Put c for this cell
                            
                            if(solve(board))
                                return true; //If it‘s the solution return true
                            else
                                board[i][j] = ‘.‘; //Otherwise go back
                        }
                    }
                    
                    return false; // 如果在某一層遍歷了1~9後還是沒有返回true則表示無解，上層填充的數字有問題，返回false到上層將相應的字符數字改回為‘.‘
                }
            }
        }
        return true;// 如果可以順利通過所有遍歷，因為每次的賦值標準是數獨合法，所以順利遍歷之後數獨也是合法的，故返回true
    }
    
    private boolean isValid(char[][] board, int row, int col, char c){
        for(int i = 0; i < 9; i++) {
            if(board[i][col] != ‘.‘ && board[i][col] == c) return false; //check row
            if(board[row][i] != ‘.‘ && board[row][i] == c) return false; //check column
            if(board[3 * (row / 3) + i / 3][ 3 * (col / 3) + i % 3] != ‘.‘ && 
board[3 * (row / 3) + i / 3][3 * (col / 3) + i % 3] == c) return false; //check 3*3 block，這裏的check 3*3 block還是很有意思的
        }
        return true;
    }
}

LeetCode解題報告—— Reverse Nodes in k-Group && Sudoku Solver