leetcode 463. Island Perimeter

You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.

Example:

[[0,1,0,0], [1,1,1,0], [0,1,0,0], [1,1,0,0]]Answer: 16Explanation: The perimeter is the 16 yellow stripes in the image below:

这个很简单，就遍历一下所有的cell呗，看它的上下左右有没有露出来，露出来就周长+1。

public class Island_Perimeter_463 {public int islandPerimeter(int[][] grid) {int zhouchang=0;int xLength=grid.length;int yLength=grid[0].length;for(int i=0;i<xLength;i++){for (int j = 0; j < yLength; j++) {if (grid[i][j] == 1) {if (i == 0 || grid[i - 1][j] == 0) {// 上zhouchang++;}if (i == xLength - 1 || grid[i + 1][j] == 0) {// 下zhouchang++;}if (j == 0 || grid[i][j - 1] == 0) {// 左zhouchang++;}if (j == yLength - 1 || grid[i][j + 1] == 0) {zhouchang++;}}}}return zhouchang;}public static void main(String[] args) {// TODO Auto-generated method stubIsland_Perimeter_463 i=new Island_Perimeter_463();int[][] a=new int[][]{{0,1,0,0},{1,1,1,0},{0,1,0,0},{1,1,0,0}};System.out.println(i.islandPerimeter(a));}}

发现大神用了DFS，用的方法真高级，这种问题我根本想不到用算法解决。

public int islandPerimeter(int[][] grid) {    if(grid.length == 0) return 0;    for(int i = 0; i < grid.length; i++){      for(int j = 0; j < grid[0].length; j++){        if(grid[i][j] == 1){          return helper(grid, i, j);        }      }      }     return 0;  }    private int helper(int[][] grid, int i, int j){    grid[i][j] = -1;    int count = 0;        if(i - 1 < 0 || grid[i - 1][j] == 0) count++;    else if(grid[i - 1][j] == 1) count += helper(grid, i - 1, j);        if(i + 1 >= grid.length || grid[i + 1][j] == 0) count++;    else if(grid[i + 1][j] == 1) count += helper(grid, i + 1, j);        if(j - 1 < 0 || grid[i][j - 1] == 0) count++;    else if(grid[i][j - 1] == 1) count += helper(grid, i, j - 1);         if(j + 1 >= grid[0].length || grid[i][j + 1] == 0) count++;    else if(grid[i][j + 1] == 1) count += helper(grid, i, j + 1);        return count;  }

他的意思就是随便找个值为1的，对它做DFS，一个个地递归下去，return的结果就是周长。