733 Flood Fill

733. Flood Fill

1. Question

Animageis represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).
Given a coordinate(sr, sc)representing the starting pixel (row and column) of the flood fill, and a pixel valuenewColor, "flood fill" the image.
To perform a "flood fill", consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.
At the end, return the modified image.
Example 1:
image = [[1,1,1],[1,1,0],[1,0,1]]
sr = 1, sc = 1, newColor = 2
Output: [[2,2,2],[2,2,0],[2,0,1]]
From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected
by a path of the same color as the starting pixel are colored with the new color.
Note the bottom corner is not colored 2, because it is not 4-directionally connected
to the starting pixel.
The length ofimageandimage[0]will be in the range[1, 50].
The given starting pixel will satisfy0 <= sr <image.lengthand0 <= sc <image[0].length.
The value of each color inimage[i][j]andnewColorwill be an integer in[0, 65535].

2. Implementation

(1) DFS
class Solution {
public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
if (image[sr][sc] != newColor) {
floodFillByDFS(image, sr, sc, image[sr][sc], newColor, directions);
return image;
public void floodFillByDFS(int[][] image, int row, int col, int color, int newColor, int[][] directions) {
image[row][col] = newColor;
for (int[] direction : directions) {
int nextRow = row + direction[0];
int nextCol = col + direction[1];
if (isValid(image, nextRow, nextCol, color)) {
floodFillByDFS(image, nextRow, nextCol, color, newColor, directions);
public boolean isValid(int[][] image, int row, int col, int color) {
return row >= 0 && row < image.length && col >= 0 && col < image[0].length && image[row][col] == color;

3. Time & Space Complexity

DFS: 时间复杂度O(n), 空间复杂度O(n)