leetcode hard模式专杀之37. Sudoku Solver
来源:互联网 发布:c json ignore 编辑:程序博客网 时间:2024/05/17 08:33
继续刷题leetcode之hard模式
这题乍一看挺吓人的 ,不过正好前几个礼拜做过八皇后问题,对回溯法有一定的记忆和认知,然后还有的技巧是位运算,hashmap记录等等,尽可能地节省空间复杂度。思路如下,先遍历一遍二维数组,本次遍历是为了收集统计数据,哪些统计数据呢?有三个,第一,每一行还差哪些数字没填, 第二,每一列还有哪些数字没填,第三,每个block还有哪些数字没填,为了节省空间,那些数字没填就交给位运算的东西去做把,理论上只需要9个位就可以标记一行或者一列有那些数字没填,这个不用我多解释了吧?已有的就用0表示,没有的就用1表示。用比特map不仅可以节省空间,更重要的是还可以通过位运算做集合的并集操作,这也是大大节省时间复杂度的。
这里有个坑要注意的是,题目的输入是char[][],也就是是char型,适当的时候要注意转换成int型,而有时又需要从int型转回char型,为了方便,我直接写了两个函数来做这种处理。
有了以上基础以后,思路大概如下,找到第一个空位置,开始尝试该位置可能的值,可能的值由以上三个可能值集合求交集而来,接下来就是回溯发来,成功就往下走,失败就返回并恢复参数,注意恢复参数很重要,不然失败的子函数就会影响到下一次递归。
代码如下:
public class Solution { public int getBlockIndex(int rowIndex, int colIndex){ return 3*(rowIndex/3)+(colIndex/3); } public int char2int(char a){ return (int)a-48; } public char int2char(int i){ return (char)(i+48); } public void solveSudoku(char[][] board) { Map<Integer,Integer> rowSet = new HashMap<>(); Map<Integer,Integer> colSet = new HashMap<>(); Map<Integer,Integer> blockSet = new HashMap<>(); // initialization for(int k = 0; k<9; k++){ rowSet.put(k, (int)Math.pow(2,9)-1); colSet.put(k, (int)Math.pow(2,9)-1); blockSet.put(k, (int)Math.pow(2,9)-1); } //after this, statistics are collected for(int row=0;row<board.length;row++){ for(int col=0;col<board[row].length;col++){ if(board[row][col]=='.'){ continue; } rowSet.put(row,rowSet.get(row) & ~(1 << char2int(board[row][col])-1)); colSet.put(col,colSet.get(col) & ~(1 << char2int(board[row][col])-1)); blockSet.put(getBlockIndex(row,col),blockSet.get(getBlockIndex(row,col)) & ~(1 << char2int(board[row][col])-1)); } } for(int row=0;row<board.length;row++) { for (int col = 0; col < board[row].length; col++) { if(board[row][col]!='.'){ continue; } int options = (rowSet.get(row) & colSet.get(col) & blockSet.get(getBlockIndex(row, col))); enumerate(board, row, col, rowSet, colSet, blockSet, options); // has to break here because enumerate can handle all, we only need to enumerate from the first '.' character break; } } } private static class Position{ int row; int col; public Position(int row, int col) { this.row = row; this.col = col; } } public Position nextPosition(char[][] board, int row, int col){ if(col<board[0].length-1){ return new Position(row, col+1); }else{ if(row<board.length-1){ return new Position(row+1, 0); }else{ return null; } } } public Position nextEmptyCel(char[][] board, int row, int col){ Position position = nextPosition(board, row, col); while(position!=null){ if(board[position.row][position.col]=='.'){ return position; } position = nextPosition(board, position.row, position.col); } return null; } public boolean enumerate(char[][] board, int row, int col, Map<Integer,Integer> rowSet, Map<Integer,Integer> colSet, Map<Integer,Integer> blockSet, int options){ if(board[row][col]!='.'){ return false; } // no option to try, means fail if(options==0){ return false; } char originBoardValue; int originRowSetValue; int originColSetValue; int originBlockSetValue; //extract positions from options for(int i = 0; i<9; i++){ if(((options >> i) & 1) ==1){ int pos = i+1; //try with value pos in row, col originBoardValue = board[row][col]; //backup board[row][col] = int2char(pos); //update rowSet originRowSetValue = rowSet.get(row); rowSet.put(row,rowSet.get(row) & ~(1 << pos-1)); //update colSet originColSetValue = colSet.get(col); colSet.put(col,colSet.get(col) & ~(1 << pos-1)); //update blockSet originBlockSetValue = blockSet.get(getBlockIndex(row, col)); blockSet.put(getBlockIndex(row, col),blockSet.get(getBlockIndex(row,col)) & ~(1 << pos-1)); //enumerate next (row,col) with '.' value, and calculate its options Position next = nextEmptyCel(board, row, col); if(next==null){ return true; } int nextRow = next.row; int nextCol = next.col; int newOptions = (rowSet.get(nextRow) & colSet.get(nextCol) & blockSet.get(getBlockIndex(nextRow, nextCol))); if(enumerate(board, nextRow,nextCol, rowSet, colSet, blockSet, newOptions)){ return true; }else{ //recover sets and continue; board[row][col]=originBoardValue; rowSet.put(row,originRowSetValue); colSet.put(col, originColSetValue); blockSet.put(getBlockIndex(row, col), originBlockSetValue); } } } return false; }}
阅读全文
0 0
- leetcode hard模式专杀之37. Sudoku Solver
- LeetCode [37. Sudoku Solver] 难度[hard]
- 37. Sudoku Solver(Hard)
- [LeetCode] 037. Sudoku Solver (Hard) (C++)
- LeetCode --- 37. Sudoku Solver
- LeetCode 37.Sudoku Solver
- [Leetcode] 37. Sudoku Solver
- [leetcode] 37. Sudoku Solver
- Leetcode 37. Sudoku Solver
- leetcode 37. Sudoku Solver
- LeetCode 37. Sudoku Solver
- leetcode.37. Sudoku Solver
- LeetCode-37.Sudoku Solver
- leetcode 37. Sudoku Solver
- (Leetcode)37. Sudoku Solver
- [LeetCode] 37. Sudoku Solver
- leetcode 37.Sudoku Solver
- leetcode 37. Sudoku Solver
- 动态网页开发基础
- Python3.x和Python2.x的区别
- 迷宫问题 POJ--3984
- DiskLruCache分析
- C3PO连接池配置及其详解
- leetcode hard模式专杀之37. Sudoku Solver
- SVM支持向量机方法——故事篇
- php 时间获取,计算,比较
- ReactNative中js与原生如何交互
- Base64 用法
- 女孩为什么都喜欢吴秀波。
- Spring Boot与Spring的区别
- Java 接收无参数post
- linux下安装搭建Memcached集群