颜色填充。编写函数，实现许多图片编辑软件都支持的“颜色填充”功能。给定一个屏幕（以二维数组表示，元素为颜色值）、一个点和一个新的颜色值，将新颜色值填入这个点的周围区域，直到原来的颜色值全都改变。
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]] 解释: 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。
说明：
- image 和 image[0] 的长度在范围 [1, 50] 内。
- 给出的初始点将满足 0 <= sr < image.length 和 0 <= sc < image[0].length。
- image[i][j] 和 newColor 表示的颜色值在范围 [0, 65535]内。
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, newColor: 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] == newColor
):
return
image[i][j] = newColor
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 newColor) {
nc = newColor;
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 newColor) {
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] == newColor) {
return;
}
image[i][j] = newColor;
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, newColor 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] == newColor {
return
}
image[i][j] = newColor
for k := 0; k < 4; k++ {
dfs(i+dirs[k], j+dirs[k+1])
}
}
dfs(sr, sc)
return image
}impl Solution {
fn dfs(i: usize, j: usize, target: i32, new_color: i32, image: &mut Vec<Vec<i32>>) {
if image[i][j] != target {
return;
}
image[i][j] = new_color;
if i != 0 {
Self::dfs(i - 1, j, target, new_color, image);
}
if j != 0 {
Self::dfs(i, j - 1, target, new_color, image);
}
if i + 1 != image.len() {
Self::dfs(i + 1, j, target, new_color, image);
}
if j + 1 != image[0].len() {
Self::dfs(i, j + 1, target, new_color, image);
}
}
pub fn flood_fill(mut image: Vec<Vec<i32>>, sr: i32, sc: i32, new_color: i32) -> Vec<Vec<i32>> {
let (sr, sc) = (sr as usize, sc as usize);
let target = image[sr][sc];
if target == new_color {
return image;
}
Self::dfs(sr, sc, target, new_color, &mut image);
image
}
}class Solution:
def floodFill(
self, image: List[List[int]], sr: int, sc: int, newColor: int
) -> List[List[int]]:
if image[sr][sc] == newColor:
return image
q = deque([(sr, sc)])
oc = image[sr][sc]
image[sr][sc] = newColor
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] = newColor
return imageclass Solution {
public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
if (image[sr][sc] == newColor) {
return image;
}
Deque<int[]> q = new ArrayDeque<>();
q.offer(new int[] {sr, sc});
int oc = image[sr][sc];
image[sr][sc] = newColor;
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] = newColor;
}
}
}
return image;
}
}class Solution {
public:
vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int newColor) {
if (image[sr][sc] == newColor) return image;
int oc = image[sr][sc];
image[sr][sc] = newColor;
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] = newColor;
}
}
}
return image;
}
};func floodFill(image [][]int, sr int, sc int, newColor int) [][]int {
if image[sr][sc] == newColor {
return image
}
oc := image[sr][sc]
q := [][]int{[]int{sr, sc}}
image[sr][sc] = newColor
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] = newColor
}
}
}
return image
}