LN : leetcode 733 Flood Fill
lc 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
imageandimage[0]will be in the range[1, 50].The given starting pixel will satisfy
0 <= sr < image.lengthand0 <= sc < image[0].length.The value of each color in
image[i][j]andnewColorwill be an integer in[0, 65535].
DFS Accepted
这道题是非常简单的dfs算法题,对于当前点,如果其原始值等于image[sr][sc]的原始值,那么将其值替换为newColor,并向其四个方向继续做dfs,直到遇到边界或者下一个原始值不等于image[sr][sc]的原始值。注意:如果image[sr][sc]的原始值就等于newColor的话,那直接返回image,不用做洪泛。
class Solution {
public:
vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int newColor) {
int color = image[sr][sc];
if (color != newColor) dfs(image, color, newColor, sr, sc);
return image;
}
void dfs(vector<vector<int>>& image, int color, int n, int sr, int sc) {
if (image[sr][sc] == color) {
image[sr][sc] = n;
if (sr - 1 >= 0) dfs(image, color, n, sr-1, sc);
if (sc - 1 >= 0) dfs(image, color, n, sr, sc-1);
if (sr + 1 < image.size()) dfs(image, color, n, sr+1, sc);
if (sc + 1 < image[0].size()) dfs(image, color, n, sr, sc+1);
}
}
};
LN : leetcode 733 Flood Fill的更多相关文章
- 【Leetcode_easy】733. Flood Fill
problem 733. Flood Fill 题意:图像处理中的泛洪填充算法,常见的有四邻域像素填充法.八邻域像素填充法.基于扫描线的像素填充法,实现方法分为递归与非递归(基于栈). 泛洪填充算法原 ...
- 【LeetCode】733. Flood Fill 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:DFS 方法二:BFS 日期 题目地址:ht ...
- [LeetCode&Python] Problem 733. Flood Fill
An image is represented by a 2-D array of integers, each integer representing the pixel value of the ...
- [LeetCode] 733. Flood Fill_Easy tag: BFS
An image is represented by a 2-D array of integers, each integer representing the pixel value of the ...
- 733. Flood Fill 简单型染色问题
[抄题]: An image is represented by a 2-D array of integers, each integer representing the pixel value ...
- 733. Flood Fill
class Solution { public: int szx,szy; vector<vector<int>> floodFill(vector<vector< ...
- Leetcode之深度优先搜索(DFS)专题-733. 图像渲染(Flood Fill)
Leetcode之深度优先搜索(DFS)专题-733. 图像渲染(Flood Fill) 深度优先搜索的解题详细介绍,点击 有一幅以二维整数数组表示的图画,每一个整数表示该图画的像素值大小,数值在 0 ...
- [LeetCode] Flood Fill 洪水填充
An image is represented by a 2-D array of integers, each integer representing the pixel value of the ...
- LeetCode刷题 Flood Fill 洪水填充问题
An image is represented by a 2-D array of integers,each integers,each integer respresenting the sta ...
随机推荐
- Vue.js 使用 Swiper.js 在 iOS 11 时出现错误
前言 在H5项目中,需要用到翻页效果,通过 Swiper 来实现,安装 Swiper npm i swiper -S 但是实际使用中,发现低版本 iOS < 11 会出现下面这个错误: Synt ...
- java如何判断字符串是否为空(小知识)
方法一: 最多人使用的一个方法, 直观, 方便, 但效率很低: if(s == null ||"".equals(s));方法二: 比较字符串长度, 效率高, 是我知道的最好一个方 ...
- Oracle:手工建库
今天学习了小布老师的手工建库视频,自己也做了一遍,下面是创建过程记录: 本地环境oracle10.2.0.1 一.前期准备工作 1.设置环境变量 [oracle@app dbs]$ vi bbk.en ...
- 深入浅出Oracle学习笔记:Undo
undo的作用是:解决oracle多用户读写一致性,以及操作可撤销或者回滚. 1.undo表空间是从10g开始进行自动管理的,几个参数如下: undo_management:回滚段手动管理还是自动管理 ...
- 【扬中集训Day6T1】 白日梦
[题目描述] 白日梦 (daydream.c/cpp/pas) 时间限制: 1 s 空间限制: 256 MB 题目描述 SR需要相当大的睡眠量 某日,他做了一个奇怪的梦,他梦见自己成为了怪物猎人,为 ...
- [laravel]phpunit
step1.install phpunit composer.json require中增加 "phpunit/phpunit":"4.0.*" 执行 comp ...
- Win7系统打开服务管理界面的几种方法汇总
转自:https://www.jb51.net/os/windows/318465.html Win7服务管理包含了计算机操作系统和应用程序提供的所有服务,但是这么多服务并非总是用户所需的.比如打印机 ...
- C++ 指针p1 p2,p1-p2 与*p1-*p2的区别
p1-p2 指 指针的地址值相减,计算两个指针之间的偏移量 *p1-*p2 指 指针指向的内存地址里面存的数值相减
- Codechef SUMCUBE
SUMCUBE code 给定无向简单图 G = (V, E)(即不存在自环和重边),以及 k = 1, 2, 或3 .求$$ \sum_{S \subseteq V} f(S)^k, $$其中 $f ...
- 1004 n^n的末位数字
题目来源: Author Ignatius.L (Hdu 1061) 基准时间限制:1 秒 空间限制:131072 KB 分值: 5 难度:1级算法题 给出一个整数N,输出N^N(N的N次方)的十 ...