733. Flood Fill

Animageis 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 valuenewColor, "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 imageand image[0]will be in the range [1, 50].

The given starting pixel will satisfy 0 <= sr < image.lengthand 0 <= sc < image[0].length.

The value of each color in image[i][j]and newColorwill be an integer in[0, 65535].

Thoughts:

1. DFS with visited map
2. DFS without visited map: only need to change the value at int[][]image.

Code: DFS with visited map

class Solution {
    public static int d [] = {0,1,0,-1,0};
    public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
        int m = image.length, n = image[0].length;
        int[][] visited = new int [m][n];
        dfs(image, sr, sc, m, n, image[sr][sc], newColor,visited);
        return image;
    }
    private void dfs(int[][] image, int sr, int sc, int m, int n, int oldColor, int newColor,int[][] visited){
        if(sr < 0 || sr >= m || sc < 0 || sc >= n || image[sr][sc] != oldColor) return;
        if (visited[sr][sc] == 1) return;
        image[sr][sc] = newColor;
        visited[sr][sc] = 1;
        for(int i = 0; i < 4; i++){
            int x = sr + d[i], y = sc + d[i + 1];
            dfs(image,x,y,m,n,oldColor, newColor,visited);
        }
    }
}

Code: DFS without visited map

class Solution {
    public static int d [] = {0,1,0,-1,0};
    public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
        int m = image.length, n = image[0].length;
        if(newColor == image[sr][sc]) return image; // only need to check if there is a change
        dfs(image, sr, sc, m, n, image[sr][sc], newColor);
        return image;
    }
    private void dfs(int[][] image, int sr, int sc, int m, int n, int oldColor, int newColor){
        if(sr < 0 || sr >= m || sc < 0 || sc >= n || image[sr][sc] != oldColor) return;
        image[sr][sc] = newColor;
        for(int i = 0; i < 4; i++){
            int x = sr + d[i], y = sc + d[i + 1];
            dfs(image,x,y,m,n,oldColor, newColor);
        }
    }
}