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].
• T-he 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].

这个题是数据结构中的图的遍历的经典问题，是一个BFS和DFS模板题，就是在给出的位置上判断颜色和要求的颜色是否不同。如果相通，直接返回就行，如果不同，就把它变为要求的颜色，并且把和它相连接的颜色（非0）全变为要求的颜色。

python代码（BFS）：

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, newColor: int) -> List[List[int]]:
        lenr, lenc = len(image), len(image[0])
        color = image[sr][sc]
        dx = [1,-1,0,0]
        dy = [0,0,1,-1]
        if color == newColor:
            return image
        else:
            num = []
            num.append((sr, sc))
            image[sr][sc] = newColor
            while len(num):
                Q = num.pop(0)
                r, c = Q[0], Q[1]
                for i in range(4):
                    x, y = r + dx[i], c + dy[i]
                    if x >= 0 and x < lenr and y >= 0 and y < lenc and image[x][y] == color:
                        num.append((x, y))
                        image[x][y] = newColor
            return image

python代码（DFS）：

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, newColor: int) -> List[List[int]]:
        lenr, lenc = len(image), len(image[0])
        color = image[sr][sc]
        if color == newColor:
            return image
        else:
            def dfs(r, c):
                if image[r][c] == color:
                    image[r][c] = newColor
                    if r >= 1: 
                        dfs(r-1, c)
                    if r + 1 < lenr: 
                        dfs(r+1, c)
                    if c >= 1: 
                        dfs(r, c-1)
                    if c + 1 < lenc: 
                        dfs(r, c+1)
            dfs(sr,sc)
            return image

最后

这个题中，可以用 a, b = c, d这个赋值，会很方便的，emmmmmmmmmm。