Leetcode[36]-Valid Sudoku

Link:https://leetcode.com/problems/valid-sudoku/

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.

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分析：验证一个数独的合法性，从三个方向入手

• 每一行不能出现重复元素
• 每一列不能出现重复元素
• 每一个九方格不能出现重复元素

我采用的是map来保存元素的数值，然后判断map中是否有，有则返回false；参数之间的关系推导如下：


代码如下

Code(c++):

class Solution {public:    bool isValidSudoku(vector<vector<char>>& board) {        //row        for(int  i = 0; i<9; i++) {            map<int, int> mp;            for (int j = 0; j<9; j++){                if(mp.find(board[i][j])!=mp.end())                    return false;                if(board[i][j] == '.')continue;                mp[board[i][j]] = j;            }        }        //cow        for(int  j = 0; j<9; j++) {            map<int, int> mp;            for (int i = 0; i<9; i++){                if(mp.find(board[i][j])!=mp.end())                    return false;                if(board[i][j] == '.')continue;                mp[board[i][j]] = i;            }        }        //board        for(int i = 0; i < 9; i++){            map<int,int> mp;            for(int j = (i/3)*3; j < 3 + (i/3)*3 ; j++){                for(int k = (3*i)%9; k <= (3*(i+1)-1)%9; k++){                    if(mp.find(board[j][k])!=mp.end())                        return false;                    if(board[j][k] == '.')continue;                    mp[board[j][k]] = k;                }            }        }        return true;    }};