leetcode 733. Flood Fill

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].

其实就是：

从给定坐标开始，将与该点相邻  并且值也相同  的点的 值都换成newColor

class Solution(object):
    def floodFill(self, image, sr, sc, newColor):
        """
        :type image: List[List[int]]
        :type sr: int
        :type sc: int
        :type newColor: int
        :rtype: List[List[int]]
        """
        old_color= image[sr][sc]
        row = len(image)
        col = len(image[0])
        
        def dfs(img, i, j):
            if i<0 or i>=row or j<0 or j>=col or img[i][j] != old_color:
                return
            img[i][j] = newColor
            dfs(img, i+1, j)
            dfs(img, i-1, j)
            dfs(img, i, j-1)
            dfs(img, i, j+1)
            
        if old_color != newColor:
            dfs(image, sr, sc)
        return image
        

BFS:

class Solution(object):
    def floodFill(self, image, sr, sc, newColor):
        """
        :type image: List[List[int]]
        :type sr: int
        :type sc: int
        :type newColor: int
        :rtype: List[List[int]]
        """        
        old_color= image[sr][sc]
        if old_color == newColor:
            return image
        
        row = len(image)
        col = len(image[0])
        
        q = [(sr, sc)]
        image[sr][sc] = newColor       
        while q:
            q2 = []
            for i,j in q:                         
                if i-1>=0 and image[i-1][j]==old_color:
                    q2.append((i-1, j))
                    image[i-1][j] = newColor       
                if i+1<row and image[i+1][j]==old_color:
                    q2.append((i+1, j))
                    image[i+1][j] = newColor  
                if j-1>=0 and image[i][j-1]==old_color:
                    q2.append((i, j-1))
                    image[i][j-1] = newColor  
                if j+1<col and image[i][j+1]==old_color:
                    q2.append((i, j+1))
                    image[i][j+1] = newColor  
            q = q2
        return image