标签：
This problem is relatively easy. The challenge is how to get a concise code. This post shares a very elegant one which traverses each cell of board only once. The code is rewritten below.
1 class Solution { 2 public: 3 bool isValidSudoku(vector<vector<char>>& board) { 4 bool row[9][9] = {false}, col[9][9] = {false}, box[9][9] = {false}; 5 for (int i = 0; i < 9; i++) { 6 for (int j = 0; j < 9; j++) { 7 if (board[i][j] != ‘.‘) { 8 int num = board[i][j] - ‘0‘ - 1, k = i / 3 * 3 + j / 3; 9 if (row[i][num] || col[j][num] || box[k][num]) return false; 10 row[i][num] = col[j][num] = box[k][num] = true; 11 } 12 } 13 } 14 return true; 15 } 16 };
标签：
原文地址：http://www.cnblogs.com/jcliBlogger/p/4792475.html
