78.Subsets
回溯法：
class Solution {
public:
vector<vector<int>> ans;
vector<vector<int>> subsets(vector<int>& nums) {
vector<int> temp;
backtrack(0, temp, nums);
return ans;
}
void backtrack(int pos, vector<int> temp, vector<int>& nums){
ans.push_back(temp);
for(int i=pos; i<nums.size(); i++){
temp.push_back(nums[i]);
backtrack(i+1, temp, nums);
temp.pop_back();
}
return;
}
};
枚举法：
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> ans;
vector<int> temp;
int len = nums.size();
ans.push_back(temp);
for(int i=0; i<len; i++){
int anslen = ans.size();
for(int j=0; j<anslen; j++){
temp = ans[j];
temp.push_back(nums[i]);
ans.push_back(temp);
}
}
return ans;
}
};
前序遍历：
class Solution {
public:
vector<vector<int>> ans;
vector<vector<int>> subsets(vector<int>& nums) {
vector<int> temp;
preorder(0, temp, nums);
return ans;
}
void preorder(int pos, vector<int> temp, vector<int>& nums){
if(pos >= nums.size()){
ans.push_back(temp);
return;
}
preorder(pos+1, temp, nums);
temp.push_back(nums[pos]);
preorder(pos+1, temp, nums);
}
};
79.Word Search
深度优先搜索 DFS + 回溯法：
很好理解的回溯法，对就继续走下去，不对就回到之前的状态另找一条路走。
根据题目的意思按照题目的走路方式写代码就对了。
代码长了点，因为是完全按照 DFS 时可能遇到的各种情况进行分类讨论了。
class Solution {
public:
bool ans = false;
vector<vector<int>> use;
bool exist(vector<vector<char>>& board, string word) {
for(int i = 0; i < board.size(); i++){
vector<int> temp;
use.push_back(temp);
for(int j = 0; j < board[i].size(); j++){
use[i].push_back(0);
}
}
backtrack(0, 0, 0, board, word);
return ans;
}
void backtrack(int num, int pos1, int pos2, vector<vector<char>>& board, string word){
if(ans)
return;
else if(num >= word.length()){
ans = true;
return;
}
else if(num == 0){
for(int i = 0; i < board.size(); i++){
for(int j = 0; j < board[i].size(); j++){
if(board[i][j] == word[0]){
use[i][j] = 1;
backtrack(1, i, j, board, word);
use[i][j] = 0;
}
}
}
}
else{
if(pos1 > 0 && board[pos1-1][pos2] == word[num] && use[pos1-1][pos2] != 1){
use[pos1-1][pos2] = 1;
backtrack(num+1, pos1-1, pos2, board, word);
use[pos1-1][pos2] = 0;
}
if(pos1 < board.size()-1 && board[pos1+1][pos2] == word[num]
&& use[pos1+1][pos2] != 1){
use[pos1+1][pos2] = 1;
backtrack(num+1, pos1+1, pos2, board, word);
use[pos1+1][pos2] = 0;
}
if(pos2 > 0 && board[pos1][pos2-1] == word[num] && use[pos1][pos2-1] != 1){
use[pos1][pos2-1] = 1;
backtrack(num+1, pos1, pos2-1, board, word);
use[pos1][pos2-1] = 0;
}
if(pos2 < board[pos1].size()-1 && board[pos1][pos2+1] == word[num]
&& use[pos1][pos2+1] != 1){
use[pos1][pos2+1] = 1;
backtrack(num+1, pos1, pos2+1, board, word);
use[pos1][pos2+1] = 0;
}
}
return;
}
};
94.Binary Tree Inorder Traversal
二叉树的中序遍历：
送分题。
/**
* 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:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> ans;
inorder(root, ans);
return ans;
}
void inorder(TreeNode* root, vector<int>& ans){
if(root == NULL)
return;
inorder(root->left, ans);
ans.push_back(root->val);
inorder(root->right, ans);
}
};
96.Unique Binary Search Trees
数学题 + 记录 + 相乘 + 相加：
记录下从 0 到 n 之间的各个数 所可能构造出来的二叉查找树（二叉搜索树）的个数。
记录下来的原因是：之后的二叉查找树个数计算会用到之前的结果。
class Solution {
public:
int numTrees(int n) {
long table[1000];
table[0] = 1;
table[1] = 1;
for(int i=2; i<=n; i++){
table[i] = 0;
for(int j=1; j<=i; j++){
table[i] += (table[j-1] * table[i-j]);
}
}
return table[n];
}
};
98.Validate Binary Search Tree
关于二叉查找树的中序遍历：
根据二叉查找树的性质，对二叉查找树进行中序遍历可以得到一个有序的序列。
通过判断进行中序遍历得到的序列是否严格有序，可以得知这个二叉查找树是否正确（有效）。
/**
* 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 isValidBST(TreeNode* root) {
vector<int> nums;
inorder(root, nums);
int len = nums.size();
for(int i=1; i<len; i++){
if(nums[i] <= nums[i-1])
return false;
}
return true;
}
void inorder(TreeNode* root, vector<int>& nums){
if(root == NULL)
return;
inorder(root->left, nums);
nums.push_back(root->val);
inorder(root->right, nums);
}
};
