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 newColor 。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，记录初始坐标的 上下左右四个方向上 像素值与初始坐标相同的相连像素点，接着再记录这四个方向上符合条件的像素点与他们对应 四个方向上 像素值与初始坐标相同的相连像素点，……，重复该过程。将所有有记录的像素点的颜色值改为 newColor 。
最后返回 经过上色渲染后的图像 。
示例 1:
输入: image = [[1,1,1],[1,1,0],[1,0,1]]，sr = 1, sc = 1, newColor = 2 输出: [[2,2,2],[2,2,0],[2,0,1]] 解析: 在图像的正中间，(坐标(sr,sc)=(1,1)),在路径上所有符合条件的像素点的颜色都被更改成2。 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, newColor = 2 输出: [[2,2,2],[2,2,2]]
提示:
m == image.lengthn == image[i].length1 <= m, n <= 500 <= image[i][j], newColor < 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
}