LWC 60:733. Flood Fill

LWC 60:733. Flood Fill

传送门:733. Flood Fill

Problem:

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].

思路: BFS,从起点开始,遇到相同的像素点加入queue,从queue中取出每个像素点时,修改成最新的颜色。BFS注意已访问过的结点需要记录下来。

JAVA代码如下:

    public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
        int n = image.length;
        int m = image[0].length;

        int[][] dir = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };

        Queue<int[]> queue = new ArrayDeque<>();
        queue.offer(new int[] {sr, sc});

        Set<Integer> vis = new HashSet<>();
        vis.add(sr * m + sc);
        while (!queue.isEmpty()) {
            int[] now = queue.poll();
            int color = image[now[0]][now[1]];
            image[now[0]][now[1]] = newColor;
            for (int[] d : dir) {
                int nx = now[0] + d[0];
                int ny = now[1] + d[1];
                if (nx >= 0 && nx < n && ny >= 0 && ny < m && image[nx][ny] == color && !vis.contains(nx * m + ny)) {
                    queue.offer(new int[] {nx, ny});
                    vis.add(nx * m + ny);
                }
            }
        }
        return image;
    }

Python代码如下:

    def floodFill(self, image, sr, sc, newColor):
        n = len(image)
        m = len(image[0])

        dir = [[1, 0],[-1, 0],[0, 1],[0, -1]]

        queue = []
        queue.append([sr, sc])
        vis = []
        vis.append(sr * m + sc)
        while (queue != []):
            now = queue[0]
            queue.remove(now)
            color = image[now[0]][now[1]]
            image[now[0]][now[1]] = newColor
            for i in range(len(dir)):
                nx = now[0] + dir[i][0]
                ny = now[1] + dir[i][1]
                if nx >= 0 and nx < n and ny >= 0 and ny < m and image[nx][ny] == color and nx * m + ny not in vis:
                    queue.append([nx, ny])
                    vis.append(nx * m + ny)

        return image

Python改进版:

    def floodFill(self, image, sr, sc, newColor):
        n = len(image)
        m = len(image[0])

        dir = [[1, 0],[-1, 0],[0, 1],[0, -1]]

        queue = []
        queue.append([sr, sc])
        vis = set()
        vis.add((sr, sc))
        while queue:
            now = queue.pop(0)
            color = image[now[0]][now[1]]
            image[now[0]][now[1]] = newColor
            for i in range(len(dir)):
                nx = now[0] + dir[i][0]
                ny = now[1] + dir[i][1]
                if nx >= 0 and nx < n and ny >= 0 and ny < m and image[nx][ny] == color and (nx, ny) not in vis:
                    queue.append([nx, ny])
                    vis.add((nx, ny))
        return image     

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

发表于

我来说两句

0 条评论
登录 后参与评论

相关文章

来自专栏Golang语言社区

Golang memory model

Introduction The Go memory model specifies the conditions under which reads of a...

4127
来自专栏Kubernetes

kube-controller-manager最佳配置

kubernetes version: 1.6.2 Kube-controller-manager Configuration kube-controller-...

4139
来自专栏Pythonista

Golang之时间格式化,计时器

1034
来自专栏ml

POJ--Strange Way to Express Integers

Strange Way to Express Integers Time Limit: 1000MS Memory Limit: 131072K ...

2667
来自专栏专注数据中心高性能网络技术研发

[Repost]The care and maintenance of your adviser

Published online  26 January 2011This article was originally published in the jo...

29411
来自专栏HansBug's Lab

1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐

1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 Time Limit: 5 Sec  Memory Limit: 64 MB Subm...

3287
来自专栏小樱的经验随笔

2017 Multi-University Training Contest - Team 9 1005&&HDU 6165 FFF at Valentine【强联通缩点+拓扑排序】

FFF at Valentine Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/6...

2805
来自专栏沃趣科技

Oracle 12c系列(九) | 通过unplug与plug方式升级pdb数据库

对于Oracle数据库升级操作,每个版本之间的升级步骤均相似,首先升级Oracle软件,然后升级数据库内的数据字典表。

1193
来自专栏互联网大杂烩

0-1背包问题(回溯法)

在0 / 1背包问题中,需对容量为c 的背包进行装载。从n 个物品中选取装入背包的物品,每件物品i 的重量为wi ,价值为pi 。 对于可行的背包装载,背包中...

462
来自专栏ml

HDUOJ-----2399GPA

GPA Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/...

3284

扫码关注云+社区