[leetcode]695. Max Area of Island小岛最大面积
Given a non-empty 2D array grid
of 0's and 1's, 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.
Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
SRE实战 互联网时代守护先锋,助力企业售后服务体系运筹帷幄!一键直达领取阿里云限量特价优惠。Example 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]]
Given the above grid, return 6
. Note the answer is not 11, because the island must be connected 4-directionally.
Example 2:
[[0,0,0,0,0,0,0,0]]
Given the above grid, return 0
.
Note: The length of each dimension in the given grid
does not exceed 50.
Solution1: DFS
code
1 class Solution { 2 public int maxAreaOfIsland(int[][] grid) { 3 // corner case 4 int max = 0; 5 if ( grid == null || grid.length==0 || grid[0].length==0) return 0; 6 for (int i = 0; i< grid.length;i++) { 7 for (int j = 0; j<grid[i].length;j++) { 8 if (grid[i][j]== 1) { 9 int[] count = new int[1]; 10 helper(grid, i, j, count); 11 max = Math.max(max , count[0]); 12 } 13 } 14 } 15 return max; 16 } 17 18 private void helper(int[][] grid, int i, int j, int[] count) { 19 // base case 20 if ((i<0) || (j<0) || (i>=grid.length) || (j>=grid[0].length) 21 || grid[i][j] != 1 ) { 22 return; 23 } 24 // recursive rule 25 count[0]++; 26 grid[i][j] = -1; // mark visited 27 helper(grid, i+1, j, count); 28 helper(grid, i-1, j, count); 29 helper(grid, i, j+1, count); 30 helper(grid, i, j-1, count); 31 } 32 }

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