1905.统计子岛屿

地址

题目

思路

先找到 B 图中为陆地而A图中为海洋的区块。

通过洪范将B图中的陆地覆盖掉。（以为如果B图中存在A图没有的部分，那么就可以说这块陆地不是A的子岛，与其相连的土地也不是）

然后再统计B中剩余岛屿的数量

代码

/*
 * @Author: your name
 * @Date: 2021-11-05 17:34:26
 * @LastEditTime: 2021-11-06 16:49:01
 * @LastEditors: Please set LastEditors
 * @Description: In User Settings Edit
 * @FilePath: /js/算法/岛屿问题/1905. 统计子岛屿/index.js
 *//**
* @param {number[][]} grid1
* @param {number[][]} grid2
* @return {number}
*/
var countSubIslands = function (grid1, grid2) {
  let [m, n] = [grid1.length, grid1[0].length]
  let count = 0
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (grid1[i][j] == 0 && grid2[i][j] == 1) {
        dfs(grid2, i, j)
      }
    }
  }
  console.log(grid2);
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (grid2[i][j] == 1) {
        count += 1
        dfs(grid2, i, j)
      }
    }
  }
  return count
};
let dfs = (grid, i, j) => {
  let [m, n] = [grid.length, grid[0].length]
  if (i < 0 || j < 0 || i >= m || j >= n) {
    return
  }
  if (grid[i][j] == 0) {
    return
  }
  grid[i][j] = 0
  dfs(grid, i - 1, j)
  dfs(grid, i + 1, j)
  dfs(grid, i, j - 1)
  dfs(grid, i, j + 1)
}