Today’s topic is to find the largest area of the island, let’s look at the topic requirements.
给定一个包含了一些 0 和 1 的非空二维数组 grid 。
一个 岛屿 是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1
必须在水平或者竖直方向上相邻。你可以假设 grid 的四个边缘都被 0（代表水包围着。
找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为 0 。)
示例 1:
[[0,0,1,0,0,0,0,1,0,0,0,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,1,1,0,1,0,0,0,0,0,0,0,0],
[0,1,0,0,1,1,0,0,1,0,1,0,0],
[0,1,0,0,1,1,0,0,1,1,1,0,0],
[0,0,0,0,0,0,0,0,0,0,1,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,0,0,0,0,0,0,1,1,0,0,0,0]]
对于上面这个给定矩阵应返回 6。注意答案不应该是 11 ，因为岛屿只能包含水平或垂
直的四个方向的 1 。
示例 2:
[[0,0,0,0,0,0,0,0]]
对于上面这个给定的矩阵, 返回 0。
First of all, according to the requirements of the question, find the maximum continuous number of 1s that meet the conditions in the two-dimensional array. We can quickly determine that such a question is the most suitable to be solved by depth-first search. The
old rule is to find the three elements of recursion.
One, function design
遇到这种需要在二维数组中查找符合条件的因素的，直接传入 **行和列**
public void dfs(int i , int j)
2. Termination conditions
当跳出这个二维数组后终止，因为加了个染色条件，还需要多个判断
在这里插入代码片
if (i<0 || j<0 || i>=row || j>=col || input[i][j] == 0)
Three, recursion direction
这种类型的提议，判断需要移动，就直接上下左右，四个方向啦！
dfs(i+1,j);
dfs(i-1,j);
dfs(i,j+1);
dfs(i,j-1);
Paste code
在这里插入代码片
public static int getMaxArea(int[][] grids) {
if (grids.length == 0) return 0 ;
input = grids ;
row = grids.length ;
col = grids[0].length ;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (input[i][j] == 1) {
dfs(i, j);
count = 0;
}
}
}
return res ;
}
static int row , col ;
static int count , res ;
static int[][] input ;
public static void dfs(int i , int j) {
if (i<0 || j<0 || i>=row || j>=col || input[i][j] == 0) {
res = Math.max(res,count) ;
return;
}
count++ ;
input[i][j] = 0 ;
dfs(i+1,j);
dfs(i-1,j);
dfs(i,j+1);
dfs(i,j-1);
}
Finished, let’s test it out
!
November 22, 2020 20:48:34