zoukankan      html  css  js  c++  java
  • leetcode 695. Max Area of Island

    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.)

    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.

    解法1:dfs

    class Solution(object):
        def maxAreaOfIsland(self, grid):
            """
            :type grid: List[List[int]]
            :rtype: int
            """
            # reset 1 to 0 use dfs, and return the ones numer
            row = len(grid)
            col = len(grid[0])
            ans = 0
            
            def dfs(grid, i, j):            
                if i==row or j==col or i<0 or j<0 or grid[i][j]==0: return 0               
                res = 1
                grid[i][j] = 0
                res += dfs(grid, i+1, j)
                res += dfs(grid, i-1, j)
                res += dfs(grid, i, j-1)
                res += dfs(grid, i, j+1)
                return res
            
            for i in range(0, row):
                for j in range(0, col):
                    if grid[i][j] == 1:
                        ans = max(ans, dfs(grid, i, j))
            return ans
            

    解法2:BFS,必须在if里将1reset为0,否则有重复:

    class Solution(object):
        def maxAreaOfIsland(self, grid):
            """
            :type grid: List[List[int]]
            :rtype: int
            """
            # reset 1 to 0 use bfs, and return the ones numer
            row = len(grid)
            col = len(grid[0])
            ans = 0
            
            def bfs(grid, i, j):            
                assert grid[i][j] == 1
                q = [(i,j)]
                grid[i][j]=0
                res = 1
                while q:                
                    q2 = []
                    for i,j in q:                    
                        if i+1<row and grid[i+1][j]==1:
                            grid[i+1][j]=0
                            q2.append((i+1,j))
                            res += 1
                        if i-1>=0 and grid[i-1][j]==1: 
                            grid[i-1][j]=0
                            q2.append((i-1,j))                        
                            res += 1
                        if j+1<col and grid[i][j+1]==1: 
                            grid[i][j+1]=0
                            q2.append((i,j+1))                        
                            res += 1
                        if j-1>=0 and grid[i][j-1]==1: 
                            grid[i][j-1]=0
                            q2.append((i,j-1))                           
                            res += 1
                    q = q2
                return res
            
            for i in range(0, row):
                for j in range(0, col):
                    if grid[i][j] == 1:
                        ans = max(ans, bfs(grid, i, j))
            return ans
            
  • 相关阅读:
    Qt5.3.0 for android windows平台下搭建及demo(虫子的博客)
    不分享,用户很少,什么都没有,没有秒传
    Entity Framework 5.0系列之约定配置
    学习SQL关联查询
    SQL语句调优
    数组总结篇(上)
    实现同一套代码针对不同平台工程的编辑和编译
    dispatch队列
    SOCKET网络编程快速上手(一)
    Javascript判断两个日期是否相等
  • 原文地址:https://www.cnblogs.com/bonelee/p/8586921.html
Copyright © 2011-2022 走看看