90 lines
2.0 KiB
Markdown
90 lines
2.0 KiB
Markdown
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---
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layout: page
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title: Flood Fill
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last_solved: 2020-06-19
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category: union find
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leetcode_url: https://leetcode.com/problems/flood-fill/
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status: Solved
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---
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Be careful with this critical tranversal section, if you make a mistake (ie: > instead of <, than it may not visit the entire grid)
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{% highlight text %}
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if r>=1: dfs(r-1, c) # go up
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if r+1<R: dfs(r+1, c) # go down
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if c>=1: dfs(r, c-1) # go left
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if c+1<C: dfs(r, c+1) # go right
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{% endhighlight %}
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{% highlight python %}
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image = [[1,1,1],
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[1,1,0],
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[1,0,1]]
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def floodFill(image, startRow, startCol, newColor):
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R = len(image)
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C = len(image[0])
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oldColor = image[startRow][startCol]
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if oldColor == newColor: return image
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def dfs(r, c):
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print(r,c)
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if image[r][c] == oldColor: # choose whether or not to advance
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image[r][c] = newColor # process the current node
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if r>=1: dfs(r-1, c) # go up
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if r+1<R: dfs(r+1, c) # go down
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if c>=1: dfs(r, c-1) # go left
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if c+1<C: dfs(r, c+1) # go right
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dfs(startRow, startCol) # mutates image, but doesn't return image
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return image
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print(floodFill(image, 1,1,2))
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[[2, 2, 2],
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[2, 2, 0],
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[2, 0, 1]]
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{% endhighlight %}
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_________________
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More concise:
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{% highlight python %}
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def floodFill(self, image: List[List[int]], startRow: int, startCol: int, newColor: int) -> List[List[int]]:
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R = len(image)
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C = len(image[0])
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oldColor = image[startRow][startCol]
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if newColor==oldColor: return image
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def dfs(i,j, newColor):
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if i<0 or i>=R or j<0 or j>=C or image[i][j]!=oldColor: # conditions to "retrack back"
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return
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dirs = [[i-1,j],[i+1,j],[i,j-1],[i,j+1]]
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for x,y in dirs:
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image[i][j]=newColor # the processing that you want to do
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dfs(x, y, newColor)
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dfs(startRow, startCol, newColor)
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return image
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{% endhighlight %}
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![image1]()
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