给你一个大小为 m x n 的整数矩阵 isWater ，它代表了一个由 陆地 和 水域 单元格组成的地图。
- 如果
isWater[i][j] == 0，格子(i, j)是一个 陆地 格子。 - 如果
isWater[i][j] == 1，格子(i, j)是一个 水域 格子。
你需要按照如下规则给每个单元格安排高度：
- 每个格子的高度都必须是非负的。
- 如果一个格子是 水域 ，那么它的高度必须为
0。 - 任意相邻的格子高度差 至多 为
1。当两个格子在正东、南、西、北方向上相互紧挨着，就称它们为相邻的格子。（也就是说它们有一条公共边）
找到一种安排高度的方案，使得矩阵中的最高高度值 最大 。
请你返回一个大小为 m x n 的整数矩阵 height ，其中 height[i][j] 是格子 (i, j) 的高度。如果有多种解法，请返回 任意一个 。
示例 1：
输入：isWater = [[0,1],[0,0]] 输出：[[1,0],[2,1]] 解释：上图展示了给各个格子安排的高度。 蓝色格子是水域格，绿色格子是陆地格。
示例 2：
输入：isWater = [[0,0,1],[1,0,0],[0,0,0]] 输出：[[1,1,0],[0,1,1],[1,2,2]] 解释：所有安排方案中，最高可行高度为 2 。 任意安排方案中，只要最高高度为 2 且符合上述规则的，都为可行方案。
提示：
m == isWater.lengthn == isWater[i].length1 <= m, n <= 1000isWater[i][j]要么是0，要么是1。- 至少有 1 个水域格子。
根据题目描述，水域的高度必须是
最后返回结果矩阵即可。
时间复杂度 isWater 的行数和列数。
class Solution:
def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:
m, n = len(isWater), len(isWater[0])
ans = [[-1] * n for _ in range(m)]
q = deque()
for i, row in enumerate(isWater):
for j, v in enumerate(row):
if v:
q.append((i, j))
ans[i][j] = 0
while q:
i, j = q.popleft()
for a, b in pairwise((-1, 0, 1, 0, -1)):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:
ans[x][y] = ans[i][j] + 1
q.append((x, y))
return ansclass Solution {
public int[][] highestPeak(int[][] isWater) {
int m = isWater.length, n = isWater[0].length;
int[][] ans = new int[m][n];
Deque<int[]> q = new ArrayDeque<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans[i][j] = isWater[i][j] - 1;
if (ans[i][j] == 0) {
q.offer(new int[] {i, j});
}
}
}
int[] dirs = {-1, 0, 1, 0, -1};
while (!q.isEmpty()) {
var p = q.poll();
int i = p[0], j = p[1];
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
ans[x][y] = ans[i][j] + 1;
q.offer(new int[] {x, y});
}
}
}
return ans;
}
}class Solution {
public:
const int dirs[5] = {-1, 0, 1, 0, -1};
vector<vector<int>> highestPeak(vector<vector<int>>& isWater) {
int m = isWater.size(), n = isWater[0].size();
vector<vector<int>> ans(m, vector<int>(n));
queue<pair<int, int>> q;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans[i][j] = isWater[i][j] - 1;
if (ans[i][j] == 0) {
q.emplace(i, j);
}
}
}
while (!q.empty()) {
auto [i, j] = q.front();
q.pop();
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
ans[x][y] = ans[i][j] + 1;
q.emplace(x, y);
}
}
}
return ans;
}
};func highestPeak(isWater [][]int) [][]int {
m, n := len(isWater), len(isWater[0])
ans := make([][]int, m)
type pair struct{ i, j int }
q := []pair{}
for i, row := range isWater {
ans[i] = make([]int, n)
for j, v := range row {
ans[i][j] = v - 1
if v == 1 {
q = append(q, pair{i, j})
}
}
}
dirs := []int{-1, 0, 1, 0, -1}
for len(q) > 0 {
p := q[0]
q = q[1:]
i, j := p.i, p.j
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 {
ans[x][y] = ans[i][j] + 1
q = append(q, pair{x, y})
}
}
}
return ans
}function highestPeak(isWater: number[][]): number[][] {
const m = isWater.length;
const n = isWater[0].length;
let ans: number[][] = [];
let q: number[][] = [];
for (let i = 0; i < m; ++i) {
ans.push(new Array(n).fill(-1));
for (let j = 0; j < n; ++j) {
if (isWater[i][j]) {
q.push([i, j]);
ans[i][j] = 0;
}
}
}
const dirs = [-1, 0, 1, 0, -1];
while (q.length) {
let tq: number[][] = [];
for (const [i, j] of q) {
for (let k = 0; k < 4; k++) {
const [x, y] = [i + dirs[k], j + dirs[k + 1]];
if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
tq.push([x, y]);
ans[x][y] = ans[i][j] + 1;
}
}
}
q = tq;
}
return ans;
}use std::collections::VecDeque;
impl Solution {
#[allow(dead_code)]
pub fn highest_peak(is_water: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
let n = is_water.len();
let m = is_water[0].len();
let mut ret_vec = vec![vec![-1; m]; n];
let mut q: VecDeque<(usize, usize)> = VecDeque::new();
let vis_pair: Vec<(i32, i32)> = vec![(-1, 0), (1, 0), (0, -1), (0, 1)];
// Initialize the return vector
for i in 0..n {
for j in 0..m {
if is_water[i][j] == 1 {
// This cell is water, the height of which must be 0
ret_vec[i][j] = 0;
q.push_back((i, j));
}
}
}
while !q.is_empty() {
// Get the front X-Y Coordinates
let (x, y) = q.front().unwrap().clone();
q.pop_front();
// Traverse through the vis pair
for d in &vis_pair {
let (dx, dy) = *d;
if Self::check_bounds((x as i32) + dx, (y as i32) + dy, n as i32, m as i32) {
if ret_vec[((x as i32) + dx) as usize][((y as i32) + dy) as usize] == -1 {
// This cell hasn't been visited, update its height
ret_vec[((x as i32) + dx) as usize][((y as i32) + dy) as usize] =
ret_vec[x][y] + 1;
// Enqueue the current cell
q.push_back((((x as i32) + dx) as usize, ((y as i32) + dy) as usize));
}
}
}
}
ret_vec
}
#[allow(dead_code)]
fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
i >= 0 && i < n && j >= 0 && j < m
}
}class Solution:
def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:
m, n = len(isWater), len(isWater[0])
ans = [[-1] * n for _ in range(m)]
q = deque()
for i, row in enumerate(isWater):
for j, v in enumerate(row):
if v:
q.append((i, j))
ans[i][j] = 0
while q:
for _ in range(len(q)):
i, j = q.popleft()
for a, b in pairwise((-1, 0, 1, 0, -1)):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:
ans[x][y] = ans[i][j] + 1
q.append((x, y))
return ansclass Solution {
public int[][] highestPeak(int[][] isWater) {
int m = isWater.length, n = isWater[0].length;
int[][] ans = new int[m][n];
Deque<int[]> q = new ArrayDeque<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans[i][j] = isWater[i][j] - 1;
if (ans[i][j] == 0) {
q.offer(new int[] {i, j});
}
}
}
int[] dirs = {-1, 0, 1, 0, -1};
while (!q.isEmpty()) {
for (int t = q.size(); t > 0; --t) {
var p = q.poll();
int i = p[0], j = p[1];
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
ans[x][y] = ans[i][j] + 1;
q.offer(new int[] {x, y});
}
}
}
}
return ans;
}
}class Solution {
public:
const int dirs[5] = {-1, 0, 1, 0, -1};
vector<vector<int>> highestPeak(vector<vector<int>>& isWater) {
int m = isWater.size(), n = isWater[0].size();
vector<vector<int>> ans(m, vector<int>(n));
queue<pair<int, int>> q;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans[i][j] = isWater[i][j] - 1;
if (ans[i][j] == 0) {
q.emplace(i, j);
}
}
}
while (!q.empty()) {
for (int t = q.size(); t; --t) {
auto [i, j] = q.front();
q.pop();
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
ans[x][y] = ans[i][j] + 1;
q.emplace(x, y);
}
}
}
}
return ans;
}
};func highestPeak(isWater [][]int) [][]int {
m, n := len(isWater), len(isWater[0])
ans := make([][]int, m)
type pair struct{ i, j int }
q := []pair{}
for i, row := range isWater {
ans[i] = make([]int, n)
for j, v := range row {
ans[i][j] = v - 1
if v == 1 {
q = append(q, pair{i, j})
}
}
}
dirs := []int{-1, 0, 1, 0, -1}
for len(q) > 0 {
for t := len(q); t > 0; t-- {
p := q[0]
q = q[1:]
i, j := p.i, p.j
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 {
ans[x][y] = ans[i][j] + 1
q = append(q, pair{x, y})
}
}
}
}
return ans
}