花花酱 LeetCode 200. Number of Islands

By zxi on September 20, 2017

Problem:

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110

11010

11000

00000

Answer: 1

Example 2:

11000

00100

00011

Idea: DFS

Use DFS to find a connected component (an island) and mark all the nodes to 0.

Time complexity: O(mn)

Space complexity: O(mn)

Solution

C++

// Author: Huahua

// Time complexity: O(mn)

// Running time: 6 ms

class Solution {

public:

int numIslands(vector<vector<char>>& grid) {

if (grid.empty()) return 0;

int m = grid.size();

int n = grid[0].size();

int ans = 0;

for (int y = 0; y < m; ++y)

for (int x = 0; x < n; ++x) {

ans += grid[y][x] - '0';

dfs(grid, x, y, m, n);

}

return ans;

}

private:

void dfs(vector<vector<char>>& grid, int x, int y, int m, int n) {

if (x < 0 || y < 0 || x >= n || y >= m || grid[y][x] == '0')

return;

grid[y][x] = '0';

dfs(grid, x + 1, y, m, n);

dfs(grid, x - 1, y, m, n);

dfs(grid, x, y + 1, m, n);

dfs(grid, x, y - 1, m, n);

}

};

Java

// Author: Huahua

// Time Complexity: O(mn)

// Running time: 13 ms

class Solution {

public int numIslands(char[][] grid) {

int m = grid.length;

if (m == 0) return 0;

int n = grid[0].length;

int ans = 0;

for (int y = 0; y < m; ++y)

for (int x = 0; x < n; ++x)

if (grid[y][x] == '1') {

++ans;

dfs(grid, x, y, n, m);

}

return ans;

}

private void dfs(char[][] grid, int x, int y, int n, int m) {

if (x < 0 || y < 0 || x >= n || y >= m || grid[y][x] == '0')

return;

grid[y][x] = '0';

dfs(grid, x + 1, y, n, m);

dfs(grid, x - 1, y, n, m);

dfs(grid, x, y + 1, n, m);

dfs(grid, x, y - 1, n, m);

}

Python

"""

Author: Huahua

Time Complexity: O(mn)

Running Time: 102 ms

"""

class Solution(object):

def numIslands(self, grid):

"""

:type grid: List[List[str]]

:rtype: int

"""

m = len(grid)

if m == 0: return 0

n = len(grid[0])

ans = 0

for y in xrange(m):

for x in xrange(n):

if grid[y][x] == '1':

ans += 1

self.__dfs(grid, x, y, n, m)

return ans

def __dfs(self, grid, x, y, n, m):

if x < 0 or y < 0 or x >=n or y >= m or grid[y][x] == '0':

return

grid[y][x] = '0'

self.__dfs(grid, x + 1, y, n, m)

self.__dfs(grid, x - 1, y, n, m)

self.__dfs(grid, x, y + 1, n, m)

self.__dfs(grid, x, y - 1, n, m)

Problem:

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:     1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:     1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

bool isSameTree(TreeNode* p, TreeNode* q) {

// Both are emtry: same

if (!p && !q) return true;

// One is emtry: not the Same

if (!p || !q) return false;

// Both are not emptry, compare the root value

if (p->val != q->val) return false;

// Compare left-subtree and right-subtree recursively.

return isSameTree(p->left, q->left)

&& isSameTree(p->right, q->right);

}

};

Java

// Author: Huahua

class Solution {

public boolean isSameTree(TreeNode p, TreeNode q) {

if (p == null && q == null) return true;

if (p == null || q == null) return false;

return p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);

}

Python3

"""

Author: Huahua

"""

class Solution:

def isSameTree(self, p, q):

if not p and not q: return True

if not p or not q: return False

return all((p.val == q.val,

self.isSameTree(p.left, q.left),

self.isSameTree(p.right, q.right)))

Follow Up

花花酱 LeetCode 572. Subtree of Another Tree

Posts tagged as “easy”

花花酱 LeetCode 200. Number of Islands

Idea: DFS

Solution

C++

Java

Python

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Problem:

Example 1:

Example 2:

Example 3:

Solution: Recursion

C++

Java

Python3

Follow Up