Today I have done Number of Islands on leetcode and leetcode's October LeetCoding Challenge with cpp.

Number of Islands

Description

Given an m x n 2d grid map of '1's (land) and '0's (water), return 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:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is '0' or '1'.

Solution

simple bfs

class Solution {
  vector<vector<int>> move{{1,0},{0,1},{-1,0},{0,-1}};
public:
  int numIslands(vector<vector<char>>& grid) {
    int sz = grid.size() * grid[0].size(), width = grid[0].size(), height = grid.size(), ans = 0;
    vector<vector<bool>> visited(height, vector<bool>(width));
    queue<pair<int, int>> start_points, q;
    for(int i = 0; i < height; ++i) {
      for(int j = 0; j < width; ++j) {
        if(grid[i][j] == '1') {
          start_points.push(make_pair(i, j));
        }
      }
    }
    while(start_points.size()) {
      auto start_point = start_points.front();
      start_points.pop();
      if(visited[start_point.first][start_point.second]) {
        continue;
      }
      ans += 1;
      visited[start_point.first][start_point.second] = true;
      q.push(start_point);
      while(q.size()) {
        auto tmp = q.front();
        int row = tmp.first, col = tmp.second;
        q.pop();
        for(int i = 0; i < 4; ++i) {
          int nrow = row+move[i][0], ncol = col+move[i][1];
          if(nrow < height && nrow >=0 && ncol < width && ncol >=0 &&
            grid[nrow][ncol] == '1' && !visited[nrow][ncol]) {
            q.push(make_pair(nrow, ncol));
            visited[nrow][ncol] = true;
          }
        }
      }
    }
    return ans;
  }
};

October LeetCoding Challenge 27

Description

Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

Notice that you should not modify the linked list.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

Constraints:

• The number of the nodes in the list is in the range [0, 104].
• -105 <= Node.val <= 105
• pos is -1 or a valid index in the linked-list.

Solution

will try O(1) space when reviewing.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
    set<ListNode*> st;
public:
    ListNode *detectCycle(ListNode *head) {
        if(!head) return NULL;
        ListNode* cur = head;
        while(cur) {
            if(st.count(cur)) return cur;
            st.insert(cur);
            cur = cur->next;
        }
        return NULL;
    }
};