代码拉取完成，页面将自动刷新
<?php
require_once 'function.php';
class Solution
{
/**
* 1567. 乘积为正数的最长子数组长度
* 正负交换 的思路没想到，这里没有理清楚；所以想到dfs，但又意识到不对，因为正负数都是在线性积累的
* max是每步判断的！否则代码好复杂！
* @param Integer[] $nums
* @return Integer
*/
function getMaxLen($nums)
{
$max = $zheng = $fu = 0;//积为正数/负数 的当前（遇到0会重新开始）最长子数组（连续）的长度
foreach ($nums as $num) {
if (!$num) {
$zheng = $fu = 0;
} else {
if ($num > 0) {
$zheng++;
if ($fu) $fu++;
} else {
// 正负交换
$_fu = $zheng + 1;
$zheng = $fu ? $fu + 1 : 0;
$fu = $_fu;
}
if ($zheng > $max) $max = $zheng;
}
}
return $max;
}
/**
* 5499. 重复至少 K 次且长度为 M 的模式
* @param Integer[] $arr
* @param Integer $m
* @param Integer $k
* @return Boolean
*/
function containsPattern($arr, $m, $k)
{
# 局部优化
$count = 0;
$l = ($k - 1) * $m;
for ($i = 0, $end = count($arr) - $m; $i < $end; $i++) {
if ($arr[$i] != $arr[$i + $m]) $count = 0;
elseif (++$count == $l) return true;
}
return false;
# 思路很好，=>连续的(k-1)*m次跳跃相等，网上那么多复杂教程？ author:Heltion
$ans = $sum = 0;
for ($i = 0; $i + $m < count($arr); $i += 1) {
if ($arr[$i] == $arr[$i + $m]) $ans = max($sum += 1, $ans);
else $sum = 0;
}
return $ans >= ($k - 1) * $m;
# 此题简单（意思输入数据不多），可以用暴力法
}
/**
* 529. 扫雷游戏
* 看似很简单，但繁琐
* @param String[][] $board
* @param Integer[] $click
* @return String[][]
*/
function updateBoard(&$board, $click)
{
list($_x, $_y) = $click;//current
//boom
if ($board[$_x][$_y] == 'M') {
$board[$_x][$_y] = 'X';
return $board;
}
if ($board[$_x][$_y] != 'E') {
return $board;
}
$xShift = [-1, -1, -1, 0, 0, 1, 1, 1];
$yShift = [-1, 0, 1, -1, 1, -1, 0, 1];
$boom = 0;
foreach ($xShift as $i => $shift) {
isset($board[$_x + $shift][$_y + $yShift[$i]])
&& ($board[$_x + $shift][$_y + $yShift[$i]] == 'M'
|| $board[$_x + $shift][$_y + $yShift[$i]] == 'X')
&& $boom++;
}
// n
if ($boom) {
$board[$_x][$_y] = strval($boom);
return $board;
}
// B
$board[$_x][$_y] = 'B';
foreach ($xShift as $i => $shift) {
isset($board[$_x + $shift][$_y + $yShift[$i]])
&& $this->updateBoard($board, [$_x + $shift, $_y + $yShift[$i]]);
}
return $board;
}
/**
* 733. 图像渲染 油漆桶工具
* 读题很难（逐个字词到位）,解题容易
* @param Integer[][] $image
* @param Integer $sr
* @param Integer $sc
* @param Integer $newColor
* @return Integer[][]
*/
function floodFill($image, $sr, $sc, $newColor)
{
static $rows, $cols;
if ($image[$sr][$sc] == $newColor) return $image;
$color = $image[$sr][$sc];
$image[$sr][$sc] = $newColor;
is_null($rows) && $rows = count($image);
is_null($cols) && $cols = count($image[0]);
$sr > 0 && $image[$sr - 1][$sc] == $color && $image = $this->floodFill($image, $sr - 1, $sc, $newColor);
$sc > 0 && $image[$sr][$sc - 1] == $color && $image = $this->floodFill($image, $sr, $sc - 1, $newColor);
$sr < $rows - 1 && $image[$sr + 1][$sc] == $color && $image = $this->floodFill($image, $sr + 1, $sc, $newColor);
$sc < $cols - 1 && $image[$sr][$sc + 1] == $color && $image = $this->floodFill($image, $sr, $sc + 1, $newColor);
return $image;
$color = $image[$sr][$sc];
$image[$sr][$sc] = $newColor;
$sr > 0 && $image[$sr - 1][$sc] == $color && $image = $this->floodFill($image, $sr - 1, $sc, $newColor);
$sc > 0 && $image[$sr][$sc - 1] == $color && $image = $this->floodFill($image, $sr, $sc - 1, $newColor);
$sr < count($image) - 1 && $image[$sr + 1][$sc] == $color && $image = $this->floodFill($image, $sr + 1, $sc, $newColor);
$sc < count($image[0]) - 1 && $image[$sr][$sc + 1] == $color && $image = $this->floodFill($image, $sr, $sc + 1, $newColor);
return $image;
}
/**
* 546. 移除盒子，消消看，得最大积分
* 很难，理清楚哪里开刀都不容易，
* 看了教程，copy代码，run是对的，提交输出错误(貌似function.static有问题)
* @param Integer[] $boxes
* @return Integer
*/
function removeBoxes($boxes)
{
return $this->calcPoints($boxes, 0, count($boxes) - 1, 0);
$map = [];
$range = [];
foreach ($boxes as $i => $x) {
//isset($map[$x]) ? $map[$x]++ : $map[$x] = 1;
if (isset($map[$x])) {
$map[$x]++;
} else {
$map[$x] = 1;
$range[$x][0] = $i;
}
$range[$x][1] = $i;
}
print_r($map);
asort($map);
foreach ($map as $x => $c) {
echo "$x : $c => " . ($range[$x][1] - $range[$x][0]) . " =>{$range[$x][0]} {$range[$x][1]}\n";
}
$total = 0;
foreach ($map as $c) {
$total += $c * $c;
}
return $total;
}
/**
* 用多维数组|字符串存储数据
* @param array $path [$l, $r, $k]
* @param null $value
* @return array|int|mixed|null
*/
function kv(array $path, $value = null)
{
static $data = [];//leetcode不支持?
# 字符串方式
$key = implode('_', $path);
if (is_null($value)) return isset($data[$key]) ? $data[$key] : 0;
return $data[$key] = $value;
# 数组方式
$p = &$data;
foreach ($path as $route) {
isset($p[$route]) or $p[$route] = [];
$p = &$p[$route];
}
if (is_null($value)) return $p ? $p : 0;
return $p = $value;
}
private $kv = [];
/**
* @param array $boxes
* @param int $l
* @param int $r
* @param int $k boxes[r]右边同值个数
* @return int|mixed
*/
function calcPoints($boxes, $l, $r, $k)
{
if ($l > $r) return 0;
if (isset($this->kv["$l-$r-$k"]) && $this->kv["$l-$r-$k"] != 0) return $this->kv["$l-$r-$k"];
while ($r > $l && $boxes[$r] == $boxes[$r - 1]) {
$r--;
$k++;
}
$this->kv["$l-$r-$k"] = $this->calcPoints($boxes, $l, $r - 1, 0)
+ ($k + 1) * ($k + 1);
for ($i = $l; $i < $r; $i++) {
if ($boxes[$i] == $boxes[$r]) {
$this->kv["$l-$r-$k"] = max(
$this->kv["$l-$r-$k"],
$this->calcPoints($boxes, $l, $i, $k + 1)
+ $this->calcPoints($boxes, $i + 1, $r - 1, 0)
);
}
}
//echo "dp[$l][$r][$k]=>{$this->kv["$l-$r-$k"]}\n";
return $this->kv["$l-$r-$k"];
#
if ($l > $r) return 0;
if (($value = $this->kv([$l, $r, $k])) != 0) return $value;
while ($r > $l && $boxes[$r] == $boxes[$r - 1]) {
$r--;
$k++;
}
$this->kv([$l, $r, $k], $this->calcPoints($boxes, $l, $r - 1, 0) + ($k + 1) * ($k + 1));
for ($i = $l; $i < $r; $i++) {
if ($boxes[$i] == $boxes[$r]) {
$this->kv([$l, $r, $k], max(
$this->kv([$l, $r, $k]),
$this->calcPoints($boxes, $l, $i, $k + 1) + $this->calcPoints($boxes, $i + 1, $r - 1, 0)
));
}
}
//echo "dp[$l][$r][$k]=>{$dp[$l][$r][$k]}\n";
return $this->kv([$l, $r, $k]);
}
/**
* 43.字符串相乘
* 业务最好懂，理顺没要太多时间
* @param String $num1
* @param String $num2
* @return String
*/
function multiply($num1, $num2)
{
# 最后统一进位，快一些。。没想到可行
if (!$num1 || !$num2) return '0';//特殊情况，不然可能返回'000'
$i = $n1 = strlen($num1);
$n2 = strlen($num2);
$sum = [];
while ($i--) {//乘数
$p0 = $n1 - $i - 1 + $n2 - 1;
$j = $n2;
while ($j--) {//被乘数
$p = $p0 - $j;
isset($sum[$p]) or $sum[$p] = 0;
$sum[$p] = $num2[$j] * $num1[$i] + $sum[$p];//当前位 总值
}
// print_r($sum);
}
unset($num1, $num2, $p0, $p1, $i, $j, $n1, $n2);
$add = 0;
# 进位处理
foreach ($sum as $i => $x) {
$x += $add;
$add = intval($x / 10);
$sum[$i] = $x % 10;
}
$add && $sum[$i + 1] = $add;
return implode('', array_reverse($sum));
# 不reverse试试
$sum2 = [];
$n = count($sum);
foreach ($sum as $i => $x) {
$x += $add;
$add = intval($x / 10);
//$sum2[$n - $i] = $x % 10;#还需要array_fill 或者 ksort，多余！
array_unshift($sum2, $x % 10);
}
//$add && array_unshift($sum2, $add);
return ($add ? $add : '') . implode('', $sum2);
# 逐个连接不如批量连接
$total = '';
foreach ($sum as $x) {
$x += $add;
$add = intval($x / 10);
$total = ($x % 10) . $total;
}
$add && $total = $add . $total;
return $total;
# 每位及时进位
if (!$num1 || !$num2) return '0';//特殊情况，不然可能返回'000'
$i = $n1 = strlen($num1);
$n2 = strlen($num2);
$sum = [];
while ($i--) {//乘数
$p0 = $n1 - $i - 1 + $n2 - 1;
$j = $n2;
$add = 0;//进位
while ($j--) {//被乘数
$p = $p0 - $j;
isset($sum[$p]) or $sum[$p] = 0;
$_ = $num2[$j] * $num1[$i] + $add + $sum[$p];//当前位 总值
$sum[$p] = $_ % 10;//个位 值
$add = $_ < 10 ? 0 : intval($_ / 10);
}
$add && $sum[$p + 1] = $add;
//print_r($sum);
}
return implode('', array_reverse($sum));
}
/**
* 括号生成
* @param Integer $n
* @return String[]
*/
function generateParenthesis($n)
{
# 按括号序列的长度递归 (a)b形式，高级，没懂
if ($n === 0) return [''];
elseif ($n === 1) return ['()'];
$ans = [];
for ($i = 0; $i < $n; ++$i) {
foreach ($this->generateParenthesis($i) as $left) {
foreach ($this->generateParenthesis($n - $i - 1) as $right) {
$ans[] = '(' . $left . ')' . $right;
}
}
}
return $ans;
# 我的压栈法
$this->combine($n - 1, $n, '(');
return $this->r;
}
private $r = [];
function combine($a, $b, $str)
{
if ($a == 0) return $this->r[] = $str . str_repeat(')', $b);
$this->combine($a - 1, $b, $str . '(');
if ($a < $b) $this->combine($a, $b - 1, $str . ')');
}
/**
* 20. 有效的括号
* 这么简单且曾经学过的题目居然还做了20min，思路一开始是计数，后来才理清楚：压栈
* 反括号来了，前一个是否是对应的正括号
* @param String $s
* @return Boolean
*/
function isValid($s)
{
$map = [
')' => '(',
']' => '[',
'}' => '{',
];
$stack = [];
for ($i = 0, $n = strlen($s); $i < $n; $i++) {
if (isset($map[$s[$i]])) {
if ($map[$s[$i]] != array_pop($stack)) return false;
else continue;
}
$stack[] = $s[$i];
}
return !$stack;
# 思路简洁，不过靠函数多次去遍历处理，不快
$l = 0;
while (strlen($s) != $l) {
$l = strlen($s);
$s = str_replace(['()', '[]', '{}'], [], $s);
}
return !$s;
}
private $map = [];//记录自己的parents
// 自己的祖先是否依赖自己
function inCircle($son, $father)
{
if (isset($this->map[$father])) {
if (in_array($son, $this->map[$father])) {
return true;
}
foreach ($this->map[$father] as $parent) {
if ($this->inCircle($son, $parent)) {
return true;
}
}
}
$this->map[$son][] = $father;
return false;
}
/**
* 207. 课程表
* 开始没搞懂题目，以为是 1->2->3，去除循环依赖，看能否选出几门课
* 提交多次才明白题意，求这是不是单向图，用递归一下就搞定了
* 掐时间做题很紧张，不断提交不断错。。洗澡才能整理好思路。
* 为什么$numCourses没用？？——另外有更高效算法
* @param Integer $numCourses
* @param Integer[][] $prerequisites
* @return Boolean
* @todo 拓扑序列，又被叫做有向无环图，DirectedAcyclicGraph（DAG）。 循环停止的条件：即不存在入度为0的顶点
*/
function canFinish($numCourses, $prerequisites)
{
foreach ($prerequisites as $group) {
list($son, $father) = $group;//儿子依赖老汉
if ($this->inCircle($son, $father)) return false;
}
return true;
}
private $board, $x, $y;
private $no = [];
/**
* 130. 被围绕的区域
* 这是我稍微有感觉一点的题目，但抽象还是花了些时间
* i,j搞反了，错了3次。。。
* 学习提高：直接在原数据上改，更快！
* @param String[][] $board
* @return NULL
*/
function solve(&$board)
{
$this->x = count($board[0]);
$this->y = count($board);
for ($i = 1; $i < $this->x - 1; $i++) {
if ($board[0][$i] == 'O') {
$this->explore($board, 1, $i);
}
if ($board[$this->y - 1][$i] == 'O') {
$this->explore($board, $this->y - 2, $i);
}
}
for ($j = 1; $j < $this->y - 1; $j++) {
if ($board[$j][0] == 'O') {
$this->explore($board, $j, 1);
}
if ($board[$j][$this->x - 1] == 'O') {
$this->explore($board, $j, $this->x - 2);
}
}
p($board);
for ($i = 1; $i < $this->x - 1; $i++) {
for ($j = 1; $j < $this->y - 1; $j++) {
if ($board[$j][$i] == 'O') {
$board[$j][$i] = 'X';
} elseif (!$board[$j][$i]) {
$board[$j][$i] = 'O';
}
}
}
}
// 行、列
function explore(&$board, $j, $i)
{
if ($board[$j][$i] != 'O') return;
$board[$j][$i] = '';
$j < $this->y - 2 && $this->explore($board, $j + 1, $i);
$j > 1 && $this->explore($board, $j - 1, $i);
$i < $this->x - 2 && $this->explore($board, $j, $i + 1);
$i > 1 && $this->explore($board, $j, $i - 1);
}
// function solve(&$board)
// {
// $this->board = $board;
// $this->x = count($board[0]);
// $this->y = count($board);
// for ($i = 1; $i < $this->x - 1; $i++) {
// if ($board[0][$i] == 'O') {
// $this->explore(1, $i);
// }
// if ($board[$this->y - 1][$i] == 'O') {
// $this->explore($this->y - 2, $i);
// }
// }
// for ($j = 1; $j < $this->y - 1; $j++) {
// if ($board[$j][0] == 'O') {
// $this->explore($j, 1);
// }
// if ($board[$j][$this->x - 1] == 'O') {
// $this->explore($j, $this->x - 2);
// }
// }
// print_r($this->no);
// for ($i = 1; $i < $this->x - 1; $i++) {
// for ($j = 1; $j < $this->y - 1; $j++) {
// if (!isset($this->no[$j][$i])) {
// $board[$j][$i] = 'X';
// }
// }
// }
// }
//
// // 行、列
// function explore($j, $i)
// {
// if ($this->board[$j][$i] != 'O' || isset($this->no[$j][$i])) return;
//
// $this->no[$j][$i] = 0;
// $j < $this->y - 2 && $this->explore($j + 1, $i);
// $j > 1 && $this->explore($j - 1, $i);
// $i < $this->x - 2 && $this->explore($j, $i + 1);
// $i > 1 && $this->explore($j, $i - 1);
// }
/**
* 343. 整数拆分 最大乘积
* 归纳法排除了一些情况，没到数学法
* @param Integer $n
* @return Integer
*/
function integerBreak($n)
{
if ($n < 2 || $n > 58) return 0;
$max = 1;
for ($i = 2; $i < $n; $i++) {
$yu = $n % $i;
$per = ($n - $yu) / $i;
$_max = $yu && $per > 1 ? ($i + $yu) * pow($i, $per - 1) : pow($i, $per);
$_max2 = pow($i, $per) * $yu;
echo "$i $per $_max $_max2\n";
if ($_max > $max) $max = $_max;
else break;
}
echo '=>', $max;
return $max;
}
/**
* 3. 无重复字符的最长子串
* 很简单，最近学会了用map；但逐个unset比内置函数substr慢了
* 另外，函数max()竟然不如strlen($subStr) > $len ? strlen($subStr) : $len语句快！
* 不过速度反馈不准确，同一代码执行两次时间差异有时挺大
* @param String $s
* @return Integer
*/
function lengthOfLongestSubstring($s)
{
//定义一个临时变量
$subStr = "";
$len = 0;
//字符串长度
$strLens = strlen($s);
for ($i = 0; $i < $strLens; $i++) {
//如果在临时变量中不存在则直接存入
$ret = strpos($subStr, $s[$i]);
if ($ret === false) {
$subStr .= $s[$i];
} else {
$len = strlen($subStr) > $len ? strlen($subStr) : $len;
$subStr .= $s[$i];
//abcabcbb 比如到了第二个a之后
$subStr = substr($subStr, $ret + 1);
}
}
$len = strlen($subStr) > $len ? strlen($subStr) : $len;
return $len;
$map = [];
$i = 0;
$n = strlen($s);
$max = 0;
while ($i < $n) {
if (isset($map[$s[$i]])) {
$max = max($max, count($map));
foreach ($map as $v => $_) {
unset($map[$v]);
if ($v == $s[$i]) {
break;
}
}
}
$map[$s[$i]] = 0;
$i++;
}
return max($max, count($map));
}
/**
* 415. 字符串相加
* 这么简单的题，我整了半天。。
* - 字符串知识
* - 引用的问题
* - 判断的问题，有时为了流程简洁一点、少点判断，结果产生判断漏洞。
* @param String $num1
* @param String $num2
* @return String
*/
function addStrings($num1, $num2)
{
# 递归交换位置，代码更简洁
if (($l2 = strlen($num2)) > ($l1 = strlen($num1))) return $this->addStrings($num2, $num1);
$add = 0;
while ($l1--) {
if (!$add && $l2 <= 0) break;
$add += ($l2 > 0 ? $num2[--$l2] : 0) + $num1[$l1];
$num1[$l1] = $add % 10;
$add = $add >= 10 ? 1 : 0;
}
$add && $num1 = '1' . $num1;
return $num1;
# 指针
$l1 = strlen($num1);
$l2 = strlen($num2);
if ($l1 > $l2) {
$l1 = $l1;
$x = $l1 - $l2;
$p = &$num1;
$p2 = &$num2;
} else {
$l1 = $l2;
$x = $l2 - $l1;
$p2 = &$num1;
$p = &$num2;
}
$add = 0;
while ($l1--) {
$add += ($l1 >= $x ? $p2[$l1 - $x] : 0);
if (!$add) break;
$add += $p[$l1];
if ($add >= 10) {
$p[$l1] = $add % 10;
$add = 1;
} else {
$p[$l1] = $add;
$add = 0;
}
}
$add && $p = '1' . $p;
return $p;
}
private $maze = ["S#O", "M..", "M.T"];
private $m;
private $n;
//private $map;//STMO#.
/**
* @param String[] $maze
* @return Integer
*/
function minimalSteps($maze)
{
$this->maze = $maze;
$this->m = $m = count($maze);
$this->n = $n = strlen($maze[0]);
$map = [];
foreach ($maze as $i => $row) {
for ($j = 0; $j < $n; $j++) {
//echo $row[$j];
isset($map[$row[$j]]) or $map[$row[$j]] = [];
array_push($map[$row[$j]], [$i, $j]);
}
}
//print_r($map);
$this->map = $map;
if (isset($map['M'])) {
if (isset($map['O'])) {
//$A = $this->findStepCount($map['S'])
}
}
return $this->findStepCount($map['S'][0], $map['T'][0]);
}
function findStepCount($start, $target, $count = 0)
{
static $roadMap = [];
// no!
if ($this->maze[$start[0]][$start[1]] === '#') return -1;
// record min step
if (isset($roadMap[$start[0]][$start[1]])) {
if ($roadMap[$start[0]][$start[1]] > $count) $roadMap[$start[0]][$start[1]] = $count;
else return -1;//走转了，放弃
} else {
$roadMap[$start[0]][$start[1]] = $count;
}
// found
if ($start == $target) {
return $roadMap[$target[0]][$target[1]];
}
// explore
$try = [];
$start[0] < $this->m - 1 && $try[] = $this->findStepCount([$start[0] + 1, $start[1]], $target, $count + 1);
$start[1] < $this->n - 1 && $try[] = $this->findStepCount([$start[0], $start[1] + 1], $target, $count + 1);
$start[0] > 0 && $try[] = $this->findStepCount([$start[0] - 1, $start[1]], $target, $count + 1);
$start[1] > 0 && $try[] = $this->findStepCount([$start[0], $start[1] - 1], $target, $count + 1);
$try = array_filter($try, function ($item) {
return $item > 0;
});
return $try ? min($try) : -1;
}
/**
* 37. 解数独
* 逻辑很简单，关键怎么写这个dfs
* @param String[][] $board
* @return NULL
*/
function solveSudoku(&$board)
{
//findEasyPoint
$empty = [];
$row = $col = $grid = [];
foreach ($board as $i => $line) {
foreach ($line as $j => $v) {
if ($v != '.') {
$row[$i][$v] = 0;
$col[$j][$v] = 0;
$grid[intval($i / 3) + 3 * intval($j / 3)][$v] = 0;
} else {
$empty[] = [$i, $j];
}
}
}
//把空位依次尝试
$this->trySet($board, $row, $col, $grid, $empty, 0);
echo "循环统计：$this->count\n";
return $board;
}
public $count = 0;
// 本来以为这是暴力法费时间，没想到就是这样。。优化就是位运算，缓存possibleValues
// 我真是想多了，以为这是困难题就需要至少代码很复杂，（解题代码也复杂）
function trySet(&$board, &$row, &$col, &$grid, &$empty, $index = 0)
{
$this->count++;
list($i, $j) = $empty[$index];
$g = intval($i / 3) + 3 * intval($j / 3);
for ($k = 1; $k <= 9; $k++) {
if (isset($row[$i][$k]) || isset($col[$j][$k]) || isset($grid[$g][$k])) continue;
$board[$i][$j] = "$k";
$row[$i][$k] = 0;
$col[$j][$k] = 0;
$grid[$g][$k] = 0;
if (!isset($empty[$index + 1])) return true;
elseif ($this->trySet($board, $row, $col, $grid, $empty, $index + 1)) return true;
unset($row[$i][$k]);//回溯
unset($col[$j][$k]);
unset($grid[$g][$k]);
}
$board[$i][$j] = '.';
return false;
}
}
$board = [["5", "3", ".", ".", "7", ".", ".", ".", "."], ["6", ".", ".", "1", "9", "5", ".", ".", "."], [".", "9", "8", ".", ".", ".", ".", "6", "."], ["8", ".", ".", ".", "6", ".", ".", ".", "3"], ["4", ".", ".", "8", ".", "3", ".", ".", "1"], ["7", ".", ".", ".", "2", ".", ".", ".", "6"], [".", "6", ".", ".", ".", ".", "2", "8", "."], [".", ".", ".", "4", "1", "9", ".", ".", "5"], [".", ".", ".", ".", "8", ".", ".", "7", "9"]];
//$board = array_fill(0, 9, array_fill(0, 9, '.'));
p($board);//die;
(new Solution())->solveSudoku($board);
p($board);
die;
//$ditu = [["X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X"],["X","X","X","X","X","X","X","X","X","O","O","O","X","X","X","X","X","X","X","X"],["X","X","X","X","X","O","O","O","X","O","X","O","X","X","X","X","X","X","X","X"],["X","X","X","X","X","O","X","O","X","O","X","O","O","O","X","X","X","X","X","X"],["X","X","X","X","X","O","X","O","O","O","X","X","X","X","X","X","X","X","X","X"],["X","X","X","X","X","O","X","X","X","X","X","X","X","X","X","X","X","X","X","X"]];
//p($ditu);
//$ditu = [["X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X"],["X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X","X"],["X","X","X","X","X","O","X","X","X","X","X","X","X","X","X","X","X","X","X","X"],["X","X","X","X","X","O","X","X","X","X","X","X","X","X","X","X","X","X","X","X"],["X","X","X","X","X","O","X","X","X","X","X","X","X","X","X","X","X","X","X","X"],["X","X","X","X","X","O","X","X","X","X","X","X","X","X","X","X","X","X","X","X"]];
//p($ditu);
$ditu = [["X", "X", "X", "X"], ["X", "O", "O", "X"], ["X", "X", "O", "X"], ["X", "O", "X", "X"]];
p($ditu);
(new Solution())->solve($ditu);
p($ditu);
die;
$dilei =
[["E", "E", "E", "E", "E", "E", "E", "E"], ["E", "E", "E", "E", "E", "E", "E", "M"], ["E", "E", "M", "E", "E", "E", "E", "E"], ["M", "E", "E", "E", "E", "E", "E", "E"], ["E", "E", "E", "E", "E", "E", "E", "E"], ["E", "E", "E", "E", "E", "E", "E", "E"], ["E", "E", "E", "E", "E", "E", "E", "E"], ["E", "E", "M", "M", "E", "E", "E", "E"]];
p($dilei);
$r = (new Solution())->updateBoard($dilei, [0, 0]);
p($r);
die;
die;
$r = (new Solution())->removeBoxes([1, 2, 3, 2]);
print_r($r);
die;
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