思路
DFS(深度优先搜索)
解题过程
将每一块不相邻的水域的鱼总和计算出来,返回鱼数量最多的那一块即可。每次找到一个还未搜索的水域,利用DFS得到与其相邻水域鱼的总和,直到所有水域都被搜索
Code
class Solution {
int dir[][] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
static int vis[][];
int row;
int col;
public int findMaxFish(int[][] grid) {
int ans = 0;
row = grid.length;
col = grid[0].length;
vis = new int[row][col];
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
int t = 0;
if (grid[i][j] > 0 && vis[i][j] == 0) {
t = dfs(i, j, grid);
}
ans = Math.max(ans, t);
}
}
return ans;
}
public int dfs(int x, int y, int[][] grid) {
vis[x][y] = 1;
int num = grid[x][y];
for (int i = 0; i < 4; i++) {
int newx = x + dir[i][0];
int newy = y + dir[i][1];
if (0 <= newx && newx < row && 0 <= newy && newy < col && vis[newx][newy] == 0 && grid[newx][newy] > 0) {
num += dfs(newx, newy, grid);
}
}
return num;
}
}
作者:菜卷
链接:https://leetcode.cn/problems/maximum-number-of-fish-in-a-grid/solutions/3032527/wang-ge-tu-zhong-yu-de-zui-da-shu-mu-by-3hnay/
来源:力扣(LeetCode)
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