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
image
andimage[0]
will be in the range[1, 50]
.The given starting pixel will satisfy
0 <= sr < image.length
and0 <= sc < image[0].length
.The value of each color in
image[i][j]
andnewColor
will 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);
}
}
};
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