79. Word Search\513. Find Bottom Left Tree Value\110. Balanced Binary Tree
- Word Search
- 题目描述
- 代码实现
- Find Bottom Left Tree Value
- 题目描述
- 代码实现
- Balanced Binary Tree
- 题目描述
- 代码实现
79. Word Search
题目描述
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.
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就是在一个矩阵字符串中，判断它能不能生成目标字符串。在矩阵中生成的目标字符串必须是连接的，是水平或者是竖直方向上连接。
代码实现
法一：最开始我是用的是广度优先。广度优先有个不好的地方，就是不好区别路径。
比如在这种情况下识别路径就很困难：
["ABCE","SFES","ADEE"]
"ABCESEEEFS"我用BFS fail的代码是：
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int m = board.size(), n = m?board[0].size():0;
bool res = false;
if(!word.size()) return true;
if(!m || !n) return false;
int loc[] = {-1, 0, 1, 0, 0, 1, 0, -1};
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(board[i][j] == word[0]) {
vector<vector<bool>> visited(m, vector<bool>(n, false));
queueint , int>> que;
que.push(make_pair(i, j));
int cnt = 0;
visited[i][j] = true;
while(!que.empty() && cnt < word.size()) {
int sz = que.size();
for(int p = 0; p < sz; p++) {
int r = que.front().first, c = que.front().second;
que.pop();
if(board[r][c] == word[cnt]) {
visited[r][c] = true;
if(cnt + 1 == word.size()) return true;
for(int k = 0; k < 4; k++)
if(r + loc[k] >= 0 && r + loc[k] < m && c + loc[k+4] >= 0 && c + loc[k+4] < n && !visited[r+loc[k]][c+loc[k+4]])
que.push(make_pair(r+loc[k], c+loc[k+4]));
}
}
cnt++;
}
// if(cnt == word.size()) res = true;
}
}
}
return res;
}
};法二：是用回溯，答案是正确的，但是超时了。看来我的回溯效率不高。以下代码是有错误的，没有经过合理的剪枝。我们应该在找到答案以后，把别的过程也终止的。
class Solution {
public:
int loc[8] = {-1, 0, 1, 0, 0, 1, 0, -1};
void judge(bool &res, vector<vector<char>>& board, string word, int& cnt, vector<vector<bool>>& vis, int m, int n, int r, int c) {
if(cnt == word.size()) {
res = true; return;
}
for(int k = 0; k < 4; k++) {
if(r + loc[k] >= 0 && r + loc[k] < m && c + loc[k+4] >= 0 && c + loc[k+4] < n && !vis[r+loc[k]][c+loc[k+4]]) {
if(board[r+loc[k]][c+loc[k+4]] == word[cnt]) {
vis[r+loc[k]][c+loc[k+4]] = true; cnt++;
judge(res, board, word, cnt, vis, m, n, r + loc[k], c + loc[k+4]);
vis[r+loc[k]][c+loc[k+4]] = false; cnt--;
}
}
}
}
bool exist(vector<vector<char>>& board, string word) {
int m = board.size(), n = m?board[0].size():0;
bool res = false;
if(!word.size()) return true;
if(!m || !n) return false;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(board[i][j] == word[0]) {
vector<vector<bool>> visited(m, vector<bool>(n, false));
visited[i][j] = true;
int cnt = 1;
judge(res, board, word, cnt, visited, m, n, i, j);
}
}
}
return res;
}
};回溯的正确解法的代码为：
class Solution {
public:
int loc[8] = {-1, 0, 1, 0, 0, 1, 0, -1};
void judge(bool &res, vector<vector<char>>& board, string word, int& cnt, vector<vector<bool>>& vis, int m, int n, int r, int c) {
if(cnt == word.size()) {
res = true; return;
}
for(int k = 0; k < 4; k++) {
if(res) return;
if(r + loc[k] >= 0 && r + loc[k] < m && c + loc[k+4] >= 0 && c + loc[k+4] < n && !vis[r+loc[k]][c+loc[k+4]]) {
if(res) return;
if(board[r+loc[k]][c+loc[k+4]] == word[cnt]) {
if(res) return;
vis[r+loc[k]][c+loc[k+4]] = true; cnt++;
judge(res, board, word, cnt, vis, m, n, r + loc[k], c + loc[k+4]);
vis[r+loc[k]][c+loc[k+4]] = false; cnt--;
}
}
}
}
bool exist(vector<vector<char>>& board, string word) {
int m = board.size(), n = m?board[0].size():0;
bool res = false;
if(!word.size()) return true;
if(!m || !n) return false;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(board[i][j] == word[0]) {
vector<vector<bool>> visited(m, vector<bool>(n, false));
visited[i][j] = true;
int cnt = 1;
judge(res, board, word, cnt, visited, m, n, i, j);
}
}
}
return res;
}
};法三：是用DFS，速度比回溯慢一些。
class Solution {
public:
int loc[8] = {-1, 0, 1, 0, 0, 1, 0, -1};
__inline vector<vector<bool>> set_flag(vector<vector<bool>> vis, int i, int j) {
vis[i][j] = true;
return vis;
}
__inline void judge(bool &res, vector<vector<char>>& board, string word, int cnt, vector<vector<bool>> vis, int m, int n, int r, int c) {
if(cnt == word.size()) {
res = true; return;
}
for(int k = 0; k < 4; k++) {
if(res) return;
if(r + loc[k] >= 0 && r + loc[k] < m && c + loc[k+4] >= 0 && c + loc[k+4] < n && !vis[r+loc[k]][c+loc[k+4]]) {
if(res) return;
if(board[r+loc[k]][c+loc[k+4]] == word[cnt]) {
if(res) return;
judge(res, board, word, cnt+1, set_flag(vis, r + loc[k], c + loc[k+4]), m, n, r + loc[k], c + loc[k+4]);
}
}
}
}
bool exist(vector<vector<char>>& board, string word) {
int m = board.size(), n = m?board[0].size():0;
bool res = false;
if(!word.size()) return true;
if(!m || !n) return false;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(board[i][j] == word[0]) {
vector<vector<bool>> visited(m, vector<bool>(n, false));
visited[i][j] = true;
int cnt = 1;
judge(res, board, word, cnt, visited, m, n, i, j);
}
}
}
return res;
}
};513. Find Bottom Left Tree Value
题目描述
Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input:
2
/ \
1 3
Output:
1Example 2:
Input:
1
/ \
2 3
/ / \
4 5 6
/
7
Output:
7Note: You may assume the tree (i.e., the given root node) is not NULL.
这道题目是求一棵树最后一行最左边的元素。
我一开始计算成了最高的左子树叶子了。
开始的时候代码是：
/**
* 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:
int findBottomLeftValue(TreeNode* root) {
int res = root->val;
queue> t;
t.push(make_pair(root, true));
while(!t.empty()) {
int size = t.size();
for(int i = 0; i < size; i++) {
TreeNode *tmp = t.front().first;
bool flg = t.front().second;
t.pop();
if(!tmp->left && !tmp->right && flg) res = tmp->val;
if(tmp->right) t.push(make_pair(tmp->right, false));
if(tmp->left) t.push(make_pair(tmp->left, true));
}
}
return res;
}
}; 代码实现
AC的代码：非常简单的一道题目。
/**
* 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:
int findBottomLeftValue(TreeNode* root) {
int res = root->val;
queue t;
t.push(root);
while(!t.empty()) {
int size = t.size();
for(int i = 0; i < size; i++) {
TreeNode *tmp = t.front();
t.pop();
if(!tmp->left && !tmp->right) res = tmp->val;
if(tmp->right) t.push(make_pair(tmp->right, false));
if(tmp->left) t.push(make_pair(tmp->left, true));
}
}
return res;
}
}; 110. Balanced Binary Tree
题目描述
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
代码实现
/**
* 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:
int compareBalance(TreeNode* root) {
if(!root) return 0;
int l1 = compareBalance(root->left);
int r1 = compareBalance(root->right);
if(l1 == -1 || r1 == -1 || abs(l1 - r1) > 1) return -1;
return max(l1, r1) + 1;
}
bool isBalanced(TreeNode* root) {
return compareBalance(root) != -1;
}
};