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简单 |
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有一幅以 m x n 的二维整数数组表示的图画 image ，其中 image[i][j] 表示该图画的像素值大小。你也被给予三个整数 sr , sc 和 color 。你应该从像素 image[sr][sc] 开始对图像进行上色 填充 。
为了完成 上色工作：
- 从初始像素开始，将其颜色改为
color。 - 对初始坐标的 上下左右四个方向上 相邻且与初始像素的原始颜色同色的像素点执行相同操作。
- 通过检查与初始像素的原始颜色相同的相邻像素并修改其颜色来继续 重复 此过程。
- 当 没有 其它原始颜色的相邻像素时 停止 操作。
最后返回经过上色渲染 修改 后的图像 。
示例 1:
输入：image = [[1,1,1],[1,1,0],[1,0,1]]，sr = 1, sc = 1, color = 2
输出：[[2,2,2],[2,2,0],[2,0,1]]
解释：在图像的正中间，坐标
(sr,sc)=(1,1) （即红色像素）,在路径上所有符合条件的像素点的颜色都被更改成相同的新颜色（即蓝色像素）。注意，右下角的像素 没有 更改为2，因为它不是在上下左右四个方向上与初始点相连的像素点。
示例 2:
输入：image = [[0,0,0],[0,0,0]], sr = 0, sc = 0, color = 0
输出：[[0,0,0],[0,0,0]]
解释：初始像素已经用 0 着色，这与目标颜色相同。因此，不会对图像进行任何更改。
提示:
m == image.lengthn == image[i].length1 <= m, n <= 500 <= image[i][j], color < 2160 <= sr < m0 <= sc < n
Flood fill 算法是从一个区域中提取若干个连通的点与其他相邻区域区分开（或分别染成不同颜色）的经典算法。因为其思路类似洪水从一个区域扩散到所有能到达的区域而得名。
最简单的实现方法是采用 DFS 的递归方法，也可以采用 BFS 的迭代来实现。
时间复杂度
class Solution:
def floodFill(
self, image: List[List[int]], sr: int, sc: int, color: int
) -> List[List[int]]:
def dfs(i, j):
if (
not 0 <= i < m
or not 0 <= j < n
or image[i][j] != oc
or image[i][j] == color
):
return
image[i][j] = color
for a, b in pairwise(dirs):
dfs(i + a, j + b)
dirs = (-1, 0, 1, 0, -1)
m, n = len(image), len(image[0])
oc = image[sr][sc]
dfs(sr, sc)
return imageclass Solution {
private int[] dirs = {-1, 0, 1, 0, -1};
private int[][] image;
private int nc;
private int oc;
public int[][] floodFill(int[][] image, int sr, int sc, int color) {
nc = color;
oc = image[sr][sc];
this.image = image;
dfs(sr, sc);
return image;
}
private void dfs(int i, int j) {
if (i < 0 || i >= image.length || j < 0 || j >= image[0].length || image[i][j] != oc
|| image[i][j] == nc) {
return;
}
image[i][j] = nc;
for (int k = 0; k < 4; ++k) {
dfs(i + dirs[k], j + dirs[k + 1]);
}
}
}class Solution {
public:
vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
int m = image.size(), n = image[0].size();
int oc = image[sr][sc];
int dirs[5] = {-1, 0, 1, 0, -1};
function<void(int, int)> dfs = [&](int i, int j) {
if (i < 0 || i >= m || j < 0 || j >= n || image[i][j] != oc || image[i][j] == color) {
return;
}
image[i][j] = color;
for (int k = 0; k < 4; ++k) {
dfs(i + dirs[k], j + dirs[k + 1]);
}
};
dfs(sr, sc);
return image;
}
};func floodFill(image [][]int, sr int, sc int, color int) [][]int {
oc := image[sr][sc]
m, n := len(image), len(image[0])
dirs := []int{-1, 0, 1, 0, -1}
var dfs func(i, j int)
dfs = func(i, j int) {
if i < 0 || i >= m || j < 0 || j >= n || image[i][j] != oc || image[i][j] == color {
return
}
image[i][j] = color
for k := 0; k < 4; k++ {
dfs(i+dirs[k], j+dirs[k+1])
}
}
dfs(sr, sc)
return image
}function floodFill(image: number[][], sr: number, sc: number, newColor: number): number[][] {
const m = image.length;
const n = image[0].length;
const target = image[sr][sc];
const dfs = (i: number, j: number) => {
if (
i < 0 ||
i === m ||
j < 0 ||
j === n ||
image[i][j] !== target ||
image[i][j] === newColor
) {
return;
}
image[i][j] = newColor;
dfs(i + 1, j);
dfs(i - 1, j);
dfs(i, j + 1);
dfs(i, j - 1);
};
dfs(sr, sc);
return image;
}impl Solution {
fn dfs(image: &mut Vec<Vec<i32>>, sr: i32, sc: i32, new_color: i32, target: i32) {
if sr < 0 || sr == (image.len() as i32) || sc < 0 || sc == (image[0].len() as i32) {
return;
}
let sr = sr as usize;
let sc = sc as usize;
if sr < 0 || image[sr][sc] == new_color || image[sr][sc] != target {
return;
}
image[sr][sc] = new_color;
let sr = sr as i32;
let sc = sc as i32;
Self::dfs(image, sr + 1, sc, new_color, target);
Self::dfs(image, sr - 1, sc, new_color, target);
Self::dfs(image, sr, sc + 1, new_color, target);
Self::dfs(image, sr, sc - 1, new_color, target);
}
pub fn flood_fill(image: Vec<Vec<i32>>, sr: i32, sc: i32, new_color: i32) -> Vec<Vec<i32>> {
let target = image[sr as usize][sc as usize];
Self::dfs(&mut image, sr, sc, new_color, target);
image
}
}class Solution:
def floodFill(
self, image: List[List[int]], sr: int, sc: int, color: int
) -> List[List[int]]:
if image[sr][sc] == color:
return image
q = deque([(sr, sc)])
oc = image[sr][sc]
image[sr][sc] = color
dirs = (-1, 0, 1, 0, -1)
while q:
i, j = q.popleft()
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < len(image) and 0 <= y < len(image[0]) and image[x][y] == oc:
q.append((x, y))
image[x][y] = color
return imageclass Solution {
public int[][] floodFill(int[][] image, int sr, int sc, int color) {
if (image[sr][sc] == color) {
return image;
}
Deque<int[]> q = new ArrayDeque<>();
q.offer(new int[] {sr, sc});
int oc = image[sr][sc];
image[sr][sc] = color;
int[] dirs = {-1, 0, 1, 0, -1};
while (!q.isEmpty()) {
int[] 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 < image.length && y >= 0 && y < image[0].length
&& image[x][y] == oc) {
q.offer(new int[] {x, y});
image[x][y] = color;
}
}
}
return image;
}
}class Solution {
public:
vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
if (image[sr][sc] == color) return image;
int oc = image[sr][sc];
image[sr][sc] = color;
queue<pair<int, int>> q;
q.push({sr, sc});
int dirs[5] = {-1, 0, 1, 0, -1};
while (!q.empty()) {
auto [a, b] = q.front();
q.pop();
for (int k = 0; k < 4; ++k) {
int x = a + dirs[k];
int y = b + dirs[k + 1];
if (x >= 0 && x < image.size() && y >= 0 && y < image[0].size() && image[x][y] == oc) {
q.push({x, y});
image[x][y] = color;
}
}
}
return image;
}
};func floodFill(image [][]int, sr int, sc int, color int) [][]int {
if image[sr][sc] == color {
return image
}
oc := image[sr][sc]
q := [][]int{[]int{sr, sc}}
image[sr][sc] = color
dirs := []int{-1, 0, 1, 0, -1}
for len(q) > 0 {
p := q[0]
q = q[1:]
for k := 0; k < 4; k++ {
x, y := p[0]+dirs[k], p[1]+dirs[k+1]
if x >= 0 && x < len(image) && y >= 0 && y < len(image[0]) && image[x][y] == oc {
q = append(q, []int{x, y})
image[x][y] = color
}
}
}
return image
}