LeetCode-695. Max Area of Island

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问题描述

You are given an m x n binary matrix grid. An island is a group of 1‘s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Example :

img

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Input: grid = [[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]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 50
  • grid[i][j] is either 0 or 1.

解答

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/*
 * @Description: 695. Max Area of Island
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * @param {number[][]} grid
 * @return {number}
 */
const maxAreaOfIsland = function (grid) {
  if (grid === null || grid.length === 0) {
    return 0
  }
  let area = 0
  const row = grid.length
  const col = grid[0].length

  const dfs = (grid, i, j) => {
    if (i < 0 || i > row - 1 || j < 0 || j > col - 1 || grid[i][j] !== 1) {
      return 0
    }
    grid[i][j] = 2
    return 1 +
           dfs(grid, i - 1, j) +
           dfs(grid, i + 1, j) +
           dfs(grid, i, j - 1) +
           dfs(grid, i, j + 1)
  }

  for (let i = 0; i < row; i++) {
    for (let j = 0; j < col; j++) {
      if (grid[i][j] === 1) {
        area = Math.max(area, dfs(grid, i, j))
      }
    }
  }
  return area
}