leetCode练习（36）

题目：Valid Sudoku

难度：easy

问题描述：

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character'.'.

A partially filled sudoku which is valid.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

解题思路：

问题的关键在于约束是什么，数独是否有效有三点需要判断：

1.每一横行没有重复数字。

2.每一纵行没有重复数字。

3.每一九宫格没有重复数字。

可以实现一个函数 isnotchongfu(char[] c)来判断输入的九个数是否重复。接下来就是找到所有需要判别的字符数组即可。

具体代码如下：

public class Solution {    int[]temp=new int[9];char[]temp2=new char[9];char c;    public boolean isValidSudoku(char[][] board) {        //横向不重复        System.out.println("你好");int i,j,k=0;for(i=0;i<9;i++){if(!isnotchongfu(board[i])){return false;}}System.out.println("heng");//3*3不重复for(i=0;i<3;i++){for(j=0;j<3;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=0;i<3;i++){for(j=3;j<6;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=0;i<3;i++){for(j=6;j<9;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=3;i<6;i++){for(j=0;j<3;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=3;i<6;i++){for(j=3;j<6;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=3;i<6;i++){for(j=6;j<9;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=6;i<9;i++){for(j=0;j<3;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=6;i<9;i++){for(j=3;j<6;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;for(i=6;i<9;i++){for(j=6;j<9;j++){temp2[k++]=board[i][j];}}if(!isnotchongfu(temp2)){return false;}k=0;//纵向不重复for(i=0;i<9;i++){//反置for(j=i;j<9;j++){c=board[i][j];board[i][j]=board[j][i];board[j][i]=c;}}for(i=0;i<9;i++){if(!isnotchongfu(board[i])){return false;}}System.out.println("zong");return true;    }    public  boolean isnotchongfu(char[] nums){        for(int j=0;j<9;j++){temp[j]=0;}for(int i=0;i<9;i++){c=nums[i];if(c!='.'){if(temp[(int)(c-'1')]==0){temp[(int)(c-'1')]++;}else{return false;}}}return true;}}

对于3*3的9个数的取值，我还没有方便的方法，希望大家的指正！