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✨ feat: add 1106、1210、1222、1245、11465、1766、1781、235、2335、2336、583、621… #788

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merged 1 commit into from
Apr 27, 2024
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✨ feat: add 1106、1210、1222、1245、11465、1766、1781、235、2335、2336、583、621… #788

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79 changes: 55 additions & 24 deletions LeetCode/1101-1110/1106. 解析布尔表达式（困难）.md
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Original file line number Diff line number Diff line change
Expand Up @@ -98,33 +98,36 @@ class Solution {
}
}
```
TypeScript 代码：
```TypeScript
function parseBoolExpr(s: string): boolean {
function calc(a: string, b: string, op: string): string {
const x = a == 't', y = b == 't'
const ans = op == '|' ? x || y : x && y
return ans ? 't' : 'f'
}
const nums = new Array<string>(s.length).fill(''), ops = new Array<string>(s.length).fill('')
let idx1 = 0, idx2 = 0
for (const c of s) {
if (c == ',') continue
if (c == 't' || c == 'f') nums[idx1++] = c
if (c == '|' || c == '&' || c == '!') ops[idx2++] = c
if (c == '(') nums[idx1++] = '-'
if (c == ')') {
let op = ops[--idx2], cur = ' '
while (idx1 > 0 && nums[idx1 - 1] != '-') {
const top = nums[--idx1]
cur = cur == ' ' ? top : calc(top, cur, op)
C++ 代码：
```C++
class Solution {
public:
bool parseBoolExpr(string s) {
deque<char> nums, ops;
for (char c : s) {
if (c == ',') continue;
if (c == 't' || c == 'f') nums.push_back(c);
if (c == '|' || c == '&' || c == '!') ops.push_back(c);
if (c == '(') nums.push_back('-');
if (c == ')') {
char op = ops.back(); ops.pop_back();
char cur = ' ';
while (!nums.empty() && nums.back() != '-') {
char top = nums.back(); nums.pop_back();
cur = cur == ' ' ? top : calc(top, cur, op);
}
if (op == '!') cur = cur == 't' ? 'f' : 't';
nums.pop_back(); nums.push_back(cur);
}
if (op == '!') cur = cur == 't' ? 'f' : 't'
idx1--; nums[idx1++] = cur
}
return nums.back() == 't';
}
return nums[idx1 - 1] == 't'
}
char calc(char a, char b, char op) {
bool x = a == 't', y = b == 't';
bool ans = op == '|' ? x | y : x & y;
return ans ? 't' : 'f';
}
};
```
Python 代码：
```Python
Expand Down Expand Up @@ -155,6 +158,34 @@ class Solution:
nums.append(cur)
return nums[-1] == 't'
```
TypeScript 代码：
```TypeScript
function parseBoolExpr(s: string): boolean {
function calc(a: string, b: string, op: string): string {
const x = a == 't', y = b == 't'
const ans = op == '|' ? x || y : x && y
return ans ? 't' : 'f'
}
const nums = new Array<string>(s.length).fill(''), ops = new Array<string>(s.length).fill('')
let idx1 = 0, idx2 = 0
for (const c of s) {
if (c == ',') continue
if (c == 't' || c == 'f') nums[idx1++] = c
if (c == '|' || c == '&' || c == '!') ops[idx2++] = c
if (c == '(') nums[idx1++] = '-'
if (c == ')') {
let op = ops[--idx2], cur = ' '
while (idx1 > 0 && nums[idx1 - 1] != '-') {
const top = nums[--idx1]
cur = cur == ' ' ? top : calc(top, cur, op)
}
if (op == '!') cur = cur == 't' ? 'f' : 't'
idx1--; nums[idx1++] = cur
}
}
return nums[idx1 - 1] == 't'
}
```
* 时间复杂度：$O(n)$
* 空间复杂度：$O(n)$

Expand Down
86 changes: 83 additions & 3 deletions LeetCode/1201-1210/1210. 穿过迷宫的最少移动次数（困难）.md
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Original file line number Diff line number Diff line change
Expand Up @@ -8,7 +8,9 @@ Tag : 「BFS」

你还记得那条风靡全球的贪吃蛇吗？

我们在一个 `n*n` 的网格上构建了新的迷宫地图，蛇的长度为 `2`，也就是说它会占去两个单元格。蛇会从左上角（`(0, 0)` 和 `(0, 1)`）开始移动。我们用 `0` 表示空单元格，用 `1` 表示障碍物。蛇需要移动到迷宫的右下角（`(n-1, n-2)` 和 `(n-1, n-1)`）。
我们在一个 `n*n` 的网格上构建了新的迷宫地图，蛇的长度为 `2`，也就是说它会占去两个单元格。蛇会从左上角（`(0, 0)` 和 `(0, 1)`）开始移动。我们用 `0` 表示空单元格，用 `1` 表示障碍物。

蛇需要移动到迷宫的右下角（`(n-1, n-2)` 和 `(n-1, n-1)`）。

每次移动，蛇可以这样走：

Expand Down Expand Up @@ -75,8 +77,7 @@ Tag : 「BFS」

在得到新蛇尾位置 $(nx, ny)$ 之后，计算新蛇头的位置 $(tx, ty)$。需要确保整条蛇没有越界，没有碰到障碍物，并且旋转转移时，额外检查 $(x + 1, y + 1)$ 位置是否合法。


代码：
Java 代码：
```Java
class Solution {
int[][] dirs = new int[][]{{1,0,0},{0,1,0},{0,0,1}};
Expand Down Expand Up @@ -105,6 +106,85 @@ class Solution {
}
}
```
C++ 代码：
```C++
class Solution {
public:
int minimumMoves(vector<vector<int>>& g) {
vector<vector<int>> dirs = {{1,0,0}, {0,1,0}, {0,0,1}};
int n = g.size();
queue<vector<int>> d;
d.push({0, 0, 0, 0});
vector<vector<vector<bool>>> vis(n, vector<vector<bool>>(n, vector<bool>(2, false)));
vis[0][0][0] = true;
while (!d.empty()) {
vector<int> info = d.front();
d.pop();
int x = info[0], y = info[1], cd = info[2], step = info[3];
for (vector<int>& dir : dirs) {
int nx = x + dir[0], ny = y + dir[1], nd = cd ^ dir[2];
int tx = nd == 0 ? nx : nx + 1, ty = nd == 0 ? ny + 1 : ny;
if (nx >= n || ny >= n || tx >= n || ty >= n) continue;
if (g[nx][ny] == 1 || g[tx][ty] == 1) continue;
if (vis[nx][ny][nd]) continue;
if (cd != nd && g[x + 1][y + 1] == 1) continue;
if (nx == n - 1 && ny == n - 2 && nd == 0) return step + 1;
d.push({nx, ny, nd, step + 1});
vis[nx][ny][nd] = true;
}
}
return -1;
}
};
```
Python 代码：
```Python
class Solution:
def minimumMoves(self, g: List[List[int]]) -> int:
dirs = [(1, 0, 0), (0, 1, 0), (0, 0, 1)]
n = len(g)
d = deque([(0,0,0,0)])
vis = [[[0]*2 for _ in range(n)] for _ in range(n)]
vis[0][0][0] = 1
while d:
x, y, cd, step = d.popleft()
for dir in dirs:
nx, ny, nd = x + dir[0], y + dir[1], cd ^ dir[2]
tx, ty = nx + (nd == 1), ny + (nd == 0)
if nx >= n or ny >= n or tx >= n or ty >= n: continue
if g[nx][ny] == 1 or g[tx][ty] == 1: continue
if vis[nx][ny][nd]: continue
if cd != nd and g[x + 1][y + 1] == 1: continue
if nx == n - 1 and ny == n - 2 and nd == 0: return step + 1
d.append((nx, ny, nd, step + 1))
vis[nx][ny][nd] = 1
return -1
```
TypeScript 代码：
```TypeScript
function minimumMoves(g: number[][]): number {
const n = g.length;
const d: [number, number, number, number][] = [[0,0,0,0]];
const vis: boolean[][][] = Array.from({ length: n }, () => Array.from({ length: n }, () => [false, false]));
vis[0][0][0] = true;
const dirs: [number, number, number][] = [[1,0,0], [0,1,0], [0,0,1]];
while (d.length > 0) {
const [x, y, cd, step] = d.shift()!;
for (const dir of dirs) {
const nx = x + dir[0], ny = y + dir[1], nd = cd ^ dir[2];
const tx = nd === 0 ? nx : nx + 1, ty = nd === 0 ? ny + 1 : ny;
if (nx >= n || ny >= n || tx >= n || ty >= n) continue;
if (g[nx][ny] === 1 || g[tx][ty] === 1) continue
if (vis[nx][ny][nd]) continue;
if (cd !== nd && g[x + 1][y + 1] === 1) continue;
if (nx === n - 1 && ny === n - 2 && nd === 0) return step + 1;
d.push([nx, ny, nd, step + 1]);
vis[nx][ny][nd] = true;
}
}
return -1;
};
```
* 时间复杂度：$O(n^2)$
* 空间复杂度：$O(n^2 \times C)$，其中 $C = 2$ 代表蛇可变状态方向

Expand Down
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