python--leetcode463. 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:

题目意思还是很简单的，就是输入一个矩阵，求其中1组成的图形的边长。

本题的思路还是很简单，楼主一开始想错了导致这一题做了大概二十分钟，有点气....

思路就是每一个为1的点，看它的上下左右是否也为1，若为1则有一边重复，最后加的时候要减去重复的边。

class Solution(object):    def islandPerimeter(self, grid):        """        :type grid: List[List[int]]        :rtype: int        """        recount=0        sum=0        n=len(grid)        m=len(grid[0])        for i in range(len(grid)):            for j in range(len(grid[i])):                if grid[i][j]==1:                    sum=sum+1                    if i-1>=0:                        if grid[i-1][j]==1 :recount+=1                    if j - 1 >= 0:                        if grid[i][j-1] == 1: recount+= 1                    if i + 1 < n:                        if grid[i + 1][j] == 1: recount += 1                    if j+1 < m:                        if grid[i][1+j] == 1: recount+=1        print(sum,recount)        return sum*4-recounts=Solution()print(s.islandPerimeter([[0,1,0,0],[1,1,1,0], [0,1,0,0], [1,1,0,0]]))

另一种解法（个人觉得更复杂）：

class Solution(object):    def islandPerimeter(self, grid):        """        :type grid: List[List[int]]        :rtype: int        """        if not grid:            return 0        def sum_adjacent(i, j):            adjacent = (i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1),            res = 0            for x, y in adjacent:                if x < 0 or y < 0 or x == len(grid) or y == len(grid[0]) or grid[x][y] == 0:                    res += 1            return res        count = 0        for i in range(len(grid)):            for j in range(len(grid[0])):                if grid[i][j] == 1:                    count += sum_adjacent(i, j)        return count