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困难 |
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在一个 n x n 的整数矩阵 grid 中，每一个方格的值 grid[i][j] 表示位置 (i, j) 的平台高度。
当开始下雨时，在时间为 t 时，水池中的水位为 t 。你可以从一个平台游向四周相邻的任意一个平台，但是前提是此时水位必须同时淹没这两个平台。假定你可以瞬间移动无限距离，也就是默认在方格内部游动是不耗时的。当然，在你游泳的时候你必须待在坐标方格里面。
你从坐标方格的左上平台 (0，0) 出发。返回 你到达坐标方格的右下平台 (n-1, n-1) 所需的最少时间 。
示例 1:
输入: grid = [[0,2],[1,3]]
输出: 3
解释:
时间为0时，你位于坐标方格的位置为 (0, 0)。
此时你不能游向任意方向，因为四个相邻方向平台的高度都大于当前时间为 0 时的水位。
等时间到达 3 时，你才可以游向平台 (1, 1). 因为此时的水位是 3，坐标方格中的平台没有比水位 3 更高的，所以你可以游向坐标方格中的任意位置
示例 2:
输入: grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]] 输出: 16 解释: 最终的路线用加粗进行了标记。 我们必须等到时间为 16，此时才能保证平台 (0, 0) 和 (4, 4) 是连通的
提示:
n == grid.lengthn == grid[i].length1 <= n <= 500 <= grid[i][j] < n2grid[i][j]中每个值 均无重复
class Solution:
def swimInWater(self, grid: List[List[int]]) -> int:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
n = len(grid)
p = list(range(n * n))
hi = [0] * (n * n)
for i, row in enumerate(grid):
for j, h in enumerate(row):
hi[h] = i * n + j
for t in range(n * n):
i, j = hi[t] // n, hi[t] % n
for a, b in [(0, -1), (0, 1), (1, 0), (-1, 0)]:
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n and grid[x][y] <= t:
p[find(x * n + y)] = find(hi[t])
if find(0) == find(n * n - 1):
return t
return -1class Solution {
private int[] p;
public int swimInWater(int[][] grid) {
int n = grid.length;
p = new int[n * n];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
}
int[] hi = new int[n * n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
hi[grid[i][j]] = i * n + j;
}
}
int[] dirs = {-1, 0, 1, 0, -1};
for (int t = 0; t < n * n; ++t) {
int i = hi[t] / n;
int j = hi[t] % n;
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k];
int y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] <= t) {
p[find(x * n + y)] = find(i * n + j);
}
if (find(0) == find(n * n - 1)) {
return t;
}
}
}
return -1;
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}class Solution {
public:
vector<int> p;
int swimInWater(vector<vector<int>>& grid) {
int n = grid.size();
p.resize(n * n);
for (int i = 0; i < p.size(); ++i) p[i] = i;
vector<int> hi(n * n);
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
hi[grid[i][j]] = i * n + j;
vector<int> dirs = {-1, 0, 1, 0, -1};
for (int t = 0; t < n * n; ++t) {
int i = hi[t] / n, j = hi[t] % n;
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] <= t)
p[find(x * n + y)] = find(hi[t]);
if (find(0) == find(n * n - 1)) return t;
}
}
return -1;
}
int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};func swimInWater(grid [][]int) int {
n := len(grid)
p := make([]int, n*n)
for i := range p {
p[i] = i
}
hi := make([]int, n*n)
for i, row := range grid {
for j, h := range row {
hi[h] = i*n + j
}
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
dirs := []int{-1, 0, 1, 0, -1}
for t := 0; t < n*n; t++ {
i, j := hi[t]/n, hi[t]%n
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] <= t {
p[find(x*n+y)] = find(hi[t])
}
if find(0) == find(n*n-1) {
return t
}
}
}
return -1
}function swimInWater(grid: number[][]): number {
const m = grid.length,
n = grid[0].length;
let visited = Array.from({ length: m }, () => new Array(n).fill(false));
let ans = 0;
let stack = [[0, 0, grid[0][0]]];
const dir = [
[0, 1],
[0, -1],
[1, 0],
[-1, 0],
];
while (stack.length) {
let [i, j] = stack.shift();
ans = Math.max(grid[i][j], ans);
if (i == m - 1 && j == n - 1) break;
for (let [dx, dy] of dir) {
let x = i + dx,
y = j + dy;
if (x < m && x > -1 && y < n && y > -1 && !visited[x][y]) {
visited[x][y] = true;
stack.push([x, y, grid[x][y]]);
}
}
stack.sort((a, b) => a[2] - b[2]);
}
return ans;
}const DIR: [(i32, i32); 4] = [(-1, 0), (1, 0), (0, -1), (0, 1)];
impl Solution {
#[allow(dead_code)]
pub fn swim_in_water(grid: Vec<Vec<i32>>) -> i32 {
let n = grid.len();
let m = grid[0].len();
let mut ret_time = 0;
let mut disjoint_set: Vec<usize> = vec![0; n * m];
// Initialize the disjoint set
for i in 0..n * m {
disjoint_set[i] = i;
}
loop {
if Self::check_and_union(&grid, &mut disjoint_set, ret_time) {
break;
}
// Otherwise, keep checking
ret_time += 1;
}
ret_time
}
#[allow(dead_code)]
fn check_and_union(grid: &Vec<Vec<i32>>, d_set: &mut Vec<usize>, cur_time: i32) -> bool {
let n = grid.len();
let m = grid[0].len();
for i in 0..n {
for j in 0..m {
if grid[i][j] != cur_time {
continue;
}
// Otherwise, let's union the square with its neighbors
for (dx, dy) in DIR {
let x = dx + (i as i32);
let y = dy + (j as i32);
if Self::check_bounds(x, y, n as i32, m as i32)
&& grid[x as usize][y as usize] <= cur_time
{
Self::union(i * m + j, (x as usize) * m + (y as usize), d_set);
}
}
}
}
Self::find(0, d_set) == Self::find(n * m - 1, d_set)
}
#[allow(dead_code)]
fn find(x: usize, d_set: &mut Vec<usize>) -> usize {
if d_set[x] != x {
d_set[x] = Self::find(d_set[x], d_set);
}
d_set[x]
}
#[allow(dead_code)]
fn union(x: usize, y: usize, d_set: &mut Vec<usize>) {
let p_x = Self::find(x, d_set);
let p_y = Self::find(y, d_set);
d_set[p_x] = p_y;
}
#[allow(dead_code)]
fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
i >= 0 && i < n && j >= 0 && j < m
}
}