目录
1--汉明距离(461)
2--目标和(494)
3--把二叉搜索树转换为累加树(538)
4--二叉树的直径(543)
5--和为L的子数组(560)
6--最短无序连续子数组(581)
7--合并二叉树(617)
8--任务调度器(621)
9--回文子串(647)
10--每日温度(739)
主要思路:
按位异或,统计1的个数;
#include
#include
class Solution {
public:
int hammingDistance(int x, int y) {
int z = x ^ y; // 按位异或
int res = 0;
while(z){
if((z % 2) == 1) res++;
z = z >> 1; // 右移
}
return res;
}
};
int main(int argc, char* argv[]){
// x = 1, y = 4
int x = 1, y = 4;
Solution S1;
int res = S1.hammingDistance(x, y);
std::cout << res << std::endl;
return 0;
}
主要思路:
转化为 0-1 背包问题,一部分数值连同 target 转化为背包容量,剩余一部分数值转化为物品,求解恰好装满背包容量的方法数;dp[j] 表示背包容量为 j 时,装满背包的方法数;
状态转移方程:dp[j] += dp[j - nums[i]],其实质是:当背包已经装了nums[i]时,剩余容量为 j - nums[i],此时装满剩余容量的方法数为 dp[j - nums[i]],遍历不同的 nums[i] 将方法数相加即可;
是有点难理解。。。
#include
#include
class Solution {
public:
int findTargetSumWays(std::vector& nums, int target) {
int sum = 0;
for(int num : nums) sum += num;
if(sum < std::abs(target)) return 0; // 数组全部元素相加相减都不能构成target
if((sum + target) % 2 == 1) return 0; // 不能二等分
int bagsize = (sum + target) / 2;
std::vector dp(bagsize + 1, 0);
dp[0] = 1;
for(int i = 0; i < nums.size(); i++){ // 遍历物品
for(int j = bagsize; j >= nums[i]; j--){ // 遍历背包容量
dp[j] += dp[j - nums[i]];
}
}
return dp[bagsize];
}
};
int main(int argc, char *argv[]) {
// nums = [1, 1, 1, 1, 1], target = 3
std::vector test = {1, 1, 1, 1, 1};
int target = 3;
Solution S1;
int res = S1.findTargetSumWays(test, target);
std::cout << res << std::endl;
return 0;
}
主要思路:
二叉树递归,本题的遍历顺序应该为 右→根→左;逐层向下传递数据,同时回溯逐层向上返回数据;
#include
#include
#include
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
class Solution {
public:
TreeNode* convertBST(TreeNode* root){
dfs(root, 0);
return root;
}
int dfs(TreeNode* root, int sum){
if(root == nullptr) return sum;
// 右
int right = dfs(root->right, sum);
// 根
root->val = root->val + right;
// 左
int left = dfs(root->left, root->val);
return left;
}
};
int main(int argc, char* argv[]){
// [4, 1, 6, 0, 2, 5, 7, null, null, null, 3, null, null, null, 8]
TreeNode* Node1 = new TreeNode(4);
TreeNode* Node2 = new TreeNode(1);
TreeNode* Node3 = new TreeNode(6);
TreeNode* Node4 = new TreeNode(0);
TreeNode* Node5 = new TreeNode(2);
TreeNode* Node6 = new TreeNode(5);
TreeNode* Node7 = new TreeNode(7);
TreeNode* Node8 = new TreeNode(3);
TreeNode* Node9 = new TreeNode(8);
Node1->left = Node2;
Node1->right = Node3;
Node2->left = Node4;
Node2->right = Node5;
Node3->left = Node6;
Node3->right = Node7;
Node5->right = Node8;
Node7->right = Node9;
Solution S1;
TreeNode* res = S1.convertBST(Node1);
// 层次遍历打印
std::queue q;
q.push(res);
while(!q.empty()){
TreeNode* top = q.front();
q.pop();
if(top != nullptr){
std::cout << top->val << " ";
}
else{
std::cout << "null" << " ";
}
if(top != nullptr){
q.push(top->left);
q.push(top->right);
}
}
return 0;
}
主要思路:
递归二叉树,对于每一个节点,递归求解左子树的结点数,右子树的结点数,向上返回最大的节点树;
二叉树的直径等于某一个节点左右子树最大路径的总结点数,例如上图中[4, 2, 1, 3]可以理解为以 1 为根节点,其左子树路径的最大结点数为[4, 2],右子树路径的最大结点数为[3];
#include
#include
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
class Solution {
public:
int diameterOfBinaryTree(TreeNode* root) {
int res = 1; // 路径经过的最大结点数
dfs(root, res);
return res - 1; // res - 1 即为直径
}
int dfs(TreeNode* root, int &res){
if(root == nullptr) return 0;
int left = dfs(root->left, res);
int right = dfs(root->right, res);
res = std::max(res, left + right + 1);
return std::max(left, right) + 1;
}
};
int main(int argc, char* argv[]){
// root = [1, 2, 3, 4, 5]
TreeNode* Node1 = new TreeNode(1);
TreeNode* Node2 = new TreeNode(2);
TreeNode* Node3 = new TreeNode(3);
TreeNode* Node4 = new TreeNode(4);
TreeNode* Node5 = new TreeNode(5);
Node1->left = Node2;
Node1->right = Node3;
Node2->left = Node4;
Node2->right = Node5;
Solution S1;
int res = S1.diameterOfBinaryTree(Node1);
std::cout << res << std::endl;
return 0;
}
主要思路:
#include
#include
#include
class Solution {
public:
int subarraySum(std::vector& nums, int k){
std::unordered_map hash_map;
hash_map[0] = 1;
int pre = 0; // 前缀和
int res = 0; // 返回的数目
for(auto num : nums){
pre += num; // 截止到当前节点的累加和
if(hash_map.find(pre - k) != hash_map.end()){ // 存在hash_map[pre - k]的节点j,即从节点j开始到当前节点的累加和为k,匹配
res += hash_map[pre - k];
}
hash_map[pre]++; // 累加和为pre的节点数+1
}
return res;
}
};
int main(int argc, char* argv[]){
// nums = [1, 1, 1], k = 2
std::vector test = {1, 1, 1};
int k = 2;
Solution S1;
int res = S1.subarraySum(test, k);
std::cout << res << std::endl;
return 0;
}
主要思路:
主要思路:
经典二叉树递归;
#include
#include
#include
#include
struct TreeNode{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right){}
};
class Solution {
public:
TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
if(root1 == nullptr) return root2;
if(root2 == nullptr) return root1;
TreeNode* root = new TreeNode(root1->val + root2->val);
root->left = mergeTrees(root1->left, root2->left);
root->right = mergeTrees(root1->right, root2->right);
return root;
}
};
int main(int argc, char* argv[]){
// root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
TreeNode *Node1 = new TreeNode(1);
TreeNode *Node2 = new TreeNode(3);
TreeNode *Node3 = new TreeNode(2);
TreeNode *Node4 = new TreeNode(5);
Node1->left = Node2;
Node1->right = Node3;
Node2->left = Node4;
TreeNode *Node5 = new TreeNode(2);
TreeNode *Node6 = new TreeNode(1);
TreeNode *Node7 = new TreeNode(3);
TreeNode *Node8 = new TreeNode(4);
TreeNode *Node9 = new TreeNode(7);
Node5->left = Node6;
Node5->right = Node7;
Node6->right= Node8;
Node7->right = Node9;
Solution S1;
TreeNode *res = S1.mergeTrees(Node1, Node5);
std::queue q;
q.push(res);
while(!q.empty()){
TreeNode* cur = q.front();
q.pop();
if(cur != nullptr){
std::cout << cur->val << " ";
q.push(cur->left);
q.push(cur->right);
}
else{
std::cout << "null" << " ";
}
}
return 0;
}
主要思路:
主要思路:
主要思路: