Leetcode DFS Series
Leetcode 77. Combinations
Given two integers n and k, return all possible combinations of k numbers out of 1 … n.
For example,
If n = 4 and k = 2, a solution is:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
class Solution {
public:
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> res;
vector<int> curr;
if (n<k) return res;
helper(res, curr, 1, n, k);
return res;
}
void helper(vector<vector<int>> &res, vector<int> &curr, int pos, int n, int k) {
if (k==0){
res.push_back(curr);
return;
}
for (int i=pos; i<=n; i++) {
curr.push_back(i);
helper(res, curr, i+1, n, k-1);
curr.pop_back();
}
}
};
Leetcode 79. Word Search
Given a 2D board and a word, find if the word exists in the grid.
The word can be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.
For example,
Given board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
word = “ABCCED”, -> returns true,
word = “SEE”, -> returns true,
word = “ABCB”, -> returns false.
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
if (board.empty() || word.empty()) return false;
const int m = board.size();
const int n = board[0].size();
for (int i=0; i<m; i++)
for (int j=0; j<n; j++)
if (board[i][j]==word[0])
if (helper(board, word, i, j, 0, m, n))
return true;
return false;
}
bool helper(vector<vector<char>> &board, string word, int i, int j, int pos, int m, int n) {
if (pos==word.size()) return true;
if (i<0 || j<0 || i>=m || j>=n || board[i][j]=='@' || board[i][j]!=word[pos])
return false;
char temp = board[i][j];
board[i][j] = '@';
if (helper(board, word, i-1, j, pos+1, m, n) || helper(board, word, i+1, j, pos+1, m, n) || helper(board, word, i, j-1, pos+1, m, n) | helper(board, word, i, j+1, pos+1, m, n))
return true;
board[i][j] = temp;
return false;
}
};
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