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  • 733. Flood Fill 填充图像

     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].

图像由二维整数数组表示,每个整数代表图像的像素值(从0到65535)。

给定表示填充的开始像素(行和列)的坐标(sr,sc)和像素值newColor,“填充”图像。

为了执行“填充填充”,考虑起始像素,加上与起始像素相同颜色的起始像素,以及与这些像素4方向连接的任何像素的4方向连接的任何像素(也与颜色相同起始像素)等等。用newColor替换上述所有像素的颜色。

最后,返回修改后的图像。
  1. public class Solution {
  2. public class Node {
  3. public Node(int row,int col) {
  4. this.row = row;
  5. this.col = col;
  6. }
  7. public int row { get; set; }
  8. public int col { get; set; }
  9. }
  10. public int[,] FloodFill(int[,] image, int sr, int sc, int newColor)
  11. {
  12. var baseColor = image[sr, sc];
  13. if (baseColor == newColor) {
  14. return image;
  15. }
  16. var q = new Queue<Node>();
  17. q.Enqueue(new Node(sr,sc));
  18. while (q.Count > 0) {
  19. var node = q.Dequeue();
  20. var r = node.row;
  21. var c = node.col;
  22. if (r - 1 >= 0 && image[r - 1, c] == baseColor) {
  23. q.Enqueue(new Node(r - 1, c));
  24. }
  25. if (r + 1 < image.GetLength(0) && image[r + 1, c] == baseColor) {
  26. q.Enqueue(new Node(r + 1, c));
  27. }
  28. if (c - 1 >= 0 && image[r, c - 1] == baseColor) {
  29. q.Enqueue(new Node(r, c - 1));
  30. }
  31. if (c + 1 < image.GetLength(1) && image[r, c + 1] == baseColor) {
  32. q.Enqueue(new Node(r, c + 1));
  33. }
  34. image[r, c] = newColor;
  35. }
  36. return image;
  37. }
  38. }




来自为知笔记(Wiz)


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  • 原文地址:https://www.cnblogs.com/xiejunzhao/p/7900995.html
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