每日一题day1：LeetCode
剑指 Offer 46. 把数字翻译成字符串
剑指 Offer 12. 矩阵中的路径
剑指 Offer 26. 树的子结构
1、剑指 Offer 46. 把数字翻译成字符串
class Solution {
public:
int translateNum(int num) {
int a = 1, b = 1, ans;
string s = to_string(num);
int size = s.size();
if(size == 1) ans = 1;
else{
for(int i = 1; i < size; i++){
string temp = s.substr(i - 1, 2);
int k = stoi(temp);
if(k >= 10 && k <= 25) ans = a + b;
a = b; b = ans;
}
}
return ans;
}
};
2、剑指 Offer 12. 矩阵中的路径
class Solution {
public:
int n, m;
bool vis[205][205], ok = false;
int dir[4][2] = {
{
-1, 0}, {
1, 0}, {
0, -1}, {
0, 1}};
bool check(int x, int y){
if(x < 0 || x >= n || y < 0 || y >= m) return false;
if(vis[x][y]) return false;
return true;
}
void dfs(vector<vector<char>>& board, string word, int x, int y, int pos){
if(pos == word.size() - 1){
ok = true;
return ;
}
for(int i = 0; i < 4; i++){
int X = x + dir[i][0];
int Y = y + dir[i][1];
if(check(X, Y) && word[pos + 1] == board[X][Y] && !ok){
vis[X][Y] = true;
dfs(board, word, X, Y, pos + 1);
vis[X][Y] = false;
}
}
}
bool exist(vector<vector<char>>& board, string word) {
m = board[0].size(), n = board.size();
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(board[i][j] == word[0]){
vis[i][j] = true;
dfs(board, word, i, j, 0);
vis[i][j] = false;
if(ok) return true;
}
}
}
return false;
}
};
3、剑指 Offer 26. 树的子结构
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool dfs(TreeNode* A, TreeNode* B){
if(B == NULL) return true;
if(A == NULL) return false;
return A -> val == B -> val && dfs(A -> left, B -> left) && dfs(A -> right, B -> right);
}
bool isSubStructure(TreeNode* A, TreeNode* B) {
if(A == NULL || B == NULL) return false;
return dfs(A, B) || isSubStructure(A -> left, B) || isSubStructure(A -> right, B);
}
};
2021/2/28完结。