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 and image[0] will be in the range [1, 50].
The given starting pixel will satisfy 0 <= sr < image.length and 0 <= sc < image[0].length.
The value of each color in image[i][j] and newColor will be an integer in [0, 65535].
Algorithm
When we change the color, we will first check if the current color is same as new color. If yes, we won't do anything. Else, we will change the color of that location. After that we need to update the color at immediate top, bottom, left and right (any pixels connected 4-directionally to those pixels).
To do this we will write a recursive function, which will take an extra parameter (which is old color).
sr and sc should be greater than zero AND sr and sc should less than length of rows and columns respectively AND the color at sr sc position should not be same as newColor.
In the recursive function, if the above conditions are satisfied we will update the color at sr sc position, and then call this recursive function for immediate right (sr+1, sc), left (sr-1, sc), top(sr, sc-1) and bottom (sr, sc+1).
-K Himaanshu Shuklaa..
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 and image[0] will be in the range [1, 50].
The given starting pixel will satisfy 0 <= sr < image.length and 0 <= sc < image[0].length.
The value of each color in image[i][j] and newColor will be an integer in [0, 65535].
Algorithm
When we change the color, we will first check if the current color is same as new color. If yes, we won't do anything. Else, we will change the color of that location. After that we need to update the color at immediate top, bottom, left and right (any pixels connected 4-directionally to those pixels).
To do this we will write a recursive function, which will take an extra parameter (which is old color).
sr and sc should be greater than zero AND sr and sc should less than length of rows and columns respectively AND the color at sr sc position should not be same as newColor.
In the recursive function, if the above conditions are satisfied we will update the color at sr sc position, and then call this recursive function for immediate right (sr+1, sc), left (sr-1, sc), top(sr, sc-1) and bottom (sr, sc+1).
Java Solution
ALSO CHECK: Other LeetCode Solutions in Java
-K Himaanshu Shuklaa..
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