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题目🦀
37. 解数独
难度困难
编写一个程序,通过填充空格来解决数独问题。
数独的解法需 遵循如下规则:
- 数字
1-9在每一行只能出现一次。 - 数字
1-9在每一列只能出现一次。 - 数字
1-9在每一个以粗实线分隔的3x3宫内只能出现一次。(请参考示例图)
数独部分空格内已填入了数字,空白格用 '.' 表示。
示例 1:
输入:board = [["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"]]
输出:[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释:输入的数独如上图所示,唯一有效的解决方案如下所示:
提示:
board.length == 9board[i].length == 9board[i][j]是一位数字或者'.'- 题目数据 保证 输入数独仅有一个解
解题思路🌵
- 有出路、有死路。
- 考虑使用回溯算法。
解题步骤🌟
- 经典回溯套路
//初始化result path
const result = []
const path = []
// [1,2,3,4]
//循环
push
backtrac()
pop
//return
- 循环行和列,尝试在每个位置放置1-9,并检验合法性
- 包括行、列、3 * 3方块的合法性
- 如果合法继续循环,直到找到一个合法的解,如果不合法,则回溯状态,并继续尝试其他的可能性
源码🔥
/**
* @param {character[][]} board
* @return {void} Do not return anything, modify board in-place instead.
*/
var solveSudoku = function(board) {
function isValid(row, col, val, board) {
let len = board.length
// 行不能重复
for(let i = 0; i < len; i++) {
if(board[row][i] === val) {
return false
}
}
// 列不能重复
for(let i = 0; i < len; i++) {
if(board[i][col] === val) {
return false
}
}
let startRow = Math.floor(row / 3) * 3
let startCol = Math.floor(col / 3) * 3
for(let i = startRow; i < startRow + 3; i++) {
for(let j = startCol; j < startCol + 3; j++) {
if(board[i][j] === val) {
return false
}
}
}
return true
}
function backTracking() {
for(let row=0;row<9;row++){
for(let col=0;col<9;col++){
if(board[row][col]!=='.'){
continue
}
//从1-9选取数字能不能放入
for(let i=1;i<=9;i++){
if(isValid(row,col,`${i}`,board)){
board[row][col]=`${i}`
if(backTracking()){
return true
}
board[row][col]=`.`
}
}
return false
}
}
return true
}
backTracking(board)
};
时间复杂度:O(N^2)
空间复杂度:O(1)
结束语🌞
那么鱼鱼的LeetCode算法篇的「leetCode」37-解数独⚡️就结束了,虽然前端对算法要求没有后端高,但是算法是编程基础,程序=数据结构➕算法,所以算法这个东西没有捷径,只能多写多练,多总结,文章的目的其实很简单,就是督促自己去完成算法练习并总结和输出,菜不菜不重要,但是热爱🔥,喜欢大家能够喜欢我的短文,也希望通过文章认识更多志同道合的朋友,如果你也喜欢折腾,欢迎加我好友,一起沙雕,一起进步。
github🤖:sudongyu
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写在最后
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