[LeetCode][JavaScript]Sudoku Solver

Sudoku Solver

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

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

 

...and its solution numbers marked in red.

https://leetcode.com/problems/sudoku-solver/

 

 

---

 

 

又是一道DFS。

test case比较弱，只有6个，并且说好了都是可解的。

每到一格先看看是不是结束了或者不需要填，否则调用findCandidate方法找出在当前这一步所有可填的数字，然后开始递归。

递归的时候就按照顺序，先横着走，走到底了换一行。

每一轮递归回来都要把默认的'.'写回去，否则会影响下一轮的结果。

开了个flag记录是不是结束，如果已经遍历完就return掉，不用再找了。

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
var solveSudoku = function(board) {
    var isComplete = false;
    dfs(0, 0);

    function dfs(x, y){
        if(isComplete){
            return;
        }
        var candidates = findCandidate(x, y);
        if(x === 8 && y == 8){
            if(board[8][8] === '.'){
                 board[8][8] = candidates[0];
            }
            isComplete = true;
            return;
        }

        if(board[x][y] !== '.'){
            if(y === 8){
                dfs(x + 1, 0);
                return;
            }else{
                dfs(x, y + 1);
                return;
            }
        }
        for(var i = 0; i < candidates.length; i++){
            board[x][y] = candidates[i];
            if(y === 8){
                dfs(x + 1, 0);
            }else{
                dfs(x, y + 1);
            }
            if(!isComplete){
                board[x][y] = '.'; 
            }else{
                return;
            }
        }
    }

    function findCandidate(x, y){
        var set = new Set();
        var candidate = [];
        var cell = -1;
        //row
        for(i = 0; i < 9; i++){
            cell = board[x][i];
            if(!set.has(cell)){
                set.add(cell);
            }   
        }
        //column
        for(i = 0; i < 9; i++){
            cell = board[i][y];
            if(!set.has(cell)){
                set.add(cell);
            }  
        }
        //square
        var offsetX = parseInt(x / 3) * 3;
        var offsetY = parseInt(y / 3) * 3;
        for(i = 0; i <= 2; i++){
            for(j = 0; j <= 2; j++){
                cell = board[i + offsetX][j + offsetY];
                if(!set.has(cell)){
                    set.add(cell);
                }
            }
        }
        //find candidate
        for(i = 1; i <= 9; i++){
            if(!set.has(i + "")){
                candidate.push(i + "");
            }
        }
        return candidate;
    } 
};

 

附赠一个UT

function test(){
    var map = [
        ['5','3','.','.','7','.','.','.','.'],
        ['6','.','.','1','9','5','.','.','.'],
        ['.','9','8','.','.','.','.','6','.'],
        ['8','.','.','.','6','.','.','.','3'],
        ['4','.','.','8','.','3','.','.','1'],
        ['7','.','.','.','2','.','.','.','6'],
        ['.','6','.','.','.','.','2','8','.'],
        ['.','.','.','4','1','9','.','.','5'],
        ['.','.','.','.','8','.','.','7','9']
    ]

    solveSudoku(map);

    for(var m = 0; m < map.length; m++){
        var line = "";
        for(var n = 0; n < map[m].length; n++){
            line += (map[m][n] + ",");
        }
        console.log(line);
    }
}