Leetcode Game of Life

Leetcode Game of Life 相关代码，本题逻辑上相对简单，就是简单的迭代。但是题目要求使用in place完成，这样就得思考一下了，我的实现方案是，先行对数组进行扫描，每个位置对应的数字的高位记录邻居为“1”的个数，最低位记录当前的状态，这样可以保证得到了新的信息又不损失以前的信息，在所有位置扫描完后，只需根据邻居为“1”的个数和当前状态，直接更新下一个状态即可，相关代码以及测试如下：

#include<iostream>
#include<vector>
#include<climits>
#include<cstdlib>

using namespace std;
// In this program, we need to implement in-place, so we used the high order
// bit to record the "1" surround the board[i][j], and the lowest bit record
// the current value. After calculated these value, we update the corespond
// value.
class Solution {
public:
void gameOfLife(vector<vector<int> >& board) {
if (board.size() == 0) {
return;
}
int m = board.size();
int n = board[0].size();
for (int i = 0; i < m; i ++) {
for (int j = 0; j < n; j ++) {
setValue(i, j, m, n, board);
}
}
for (int i = 0; i < m; i ++) {
for (int j = 0; j < n; j ++) {
evaluate(i, j, board);
}
}
}

// Calculate the number of "1" in board[i][j] 8 neighbor's.
// For record these information in-place, we use the char
// high order bit to recorde the number of "1", use the
// lowest bit to record the origin value
void setValue(int i, int j, int m, int n, vector<vector<int> >& board) {
int count = 0;
for (int row = i - 1; row <= i + 1; row ++) {
for (int col = j - 1; col <= j + 1; col ++) {
if ((row >= 0 && row < m) && (col >= 0 && col < n)) {
if ((row != i || col != j) && (board[row][col] & 1) == 1) {
count ++;
}
}
}
}
count <<= 1;
board[i][j] = count + board[i][j];
}

// Decide the state of the next iteration for the pre information
// which recorded in the high order bits.
void evaluate(int i, int j, vector<vector<int> >& board) {
int current_state = board[i][j] & 1;
int count = board[i][j] >> 1;
if (current_state == 0 && count == 3) {
board[i][j] = 1;
} else if (current_state == 1) {
if (count < 2) {
board[i][j] = 0;
} else if (count > 3) {
board[i][j] = 0;
} else {
board[i][j] = current_state;
}
} else {
board[i][j] = current_state;
}
}
};

void print_cur(vector<vector<int> >& board) {
for (int i = 0; i < board.size(); i ++) {
for (int j = 0; j < board[0].size(); j ++) {
cout<<board[i][j]<<" ";
}
cout<<endl;
}
cout<<endl;
}

int main(int argc, char* argv[]) {
Solution so;
int row = atoi(argv[1]);
int col = atoi(argv[2]);
vector<vector<int> > test(row, vector<int>(col, 1));
for (int i = 0; i < row; i ++) {
for (int j = 0; j < col; j ++) {
test[i][j] = rand() % 2;
}
}
print_cur(test);
for (int time = 0; time < (row + col) * 2; time ++) {
cout<<"time: "<<time<<endl;
so.gameOfLife(test);
print_cur(test);
}
cout<<endl;
return 0;
}