733. Flood Fill - Easy
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An image
is 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 value newColor
, "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:
Input: 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]] Explanation: 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.
Note:
- The length of
image
andimage[0]
will be in the range[1, 50]
. - The given starting pixel will satisfy
0 <= sr < image.length
and0 <= sc < image[0].length
. - The value of each color in
image[i][j]
andnewColor
will be an integer in[0, 65535]
.
注意要加上check color的条件,否则stackoverflow
time: O(mn), space: O(mn) -- worst case, might need to process all nodes
class Solution { public int[][] floodFill(int[][] image, int sr, int sc, int newColor) { int oldColor = image[sr][sc]; if(oldColor != newColor) { dfs(image, sr, sc, oldColor, newColor); } return image; } public void dfs(int[][] image, int r, int c, int oldColor, int newColor) { if(r < 0 || r > image.length - 1 || c < 0 || c > image[0].length - 1) { return; } if(image[r][c] == oldColor) { image[r][c] = newColor; dfs(image, r - 1, c, oldColor, newColor); dfs(image, r, c + 1, oldColor, newColor); dfs(image, r + 1, c, oldColor, newColor); dfs(image, r, c - 1, oldColor, newColor); } } }
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