leetcode 哈希表相关算法刷题笔记
2022/1/6 20:04:57
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刷题笔记
- hash table 算法leetcode专栏
- leetcode 242 有效的字母异位词
- leetcode 383 赎金信
- leetcode 49 字母异位词分组
- leetcode 138 复制带随机指针的链表
- leetcode 1002 查找共用字符
- leetcode 349 两个数组的交集
- leetcode 350 两个数组的交集II
- leetcode 202 快乐数
- leetcode 1 两数之和
- leetcode 15 三数之和
- leetcode 18 四数之和
- leetcode 454 四数相加II
hash table 算法leetcode专栏
leetcode 242 有效的字母异位词
class Solution {
public:
bool isAnagram(string s, string t) {
vector<int> hashVec(26, 0);
for(int i = 0; i < s.size(); i++)
{
hashVec[s[i] -'a']++;
}
for(int i = 0; i < t.size(); i++)
{
hashVec[t[i] - 'a']--;
}
bool flag = true;
for(int i = 0; i < hashVec.size(); i++)
{
if(hashVec[i] != 0)
{
flag = false;
break;
}
}
return flag;
}
};
leetcode 383 赎金信
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
vector<int> hashVec(26, 0);
for(int i = 0; i < magazine.size(); i++)
{
hashVec[magazine[i] -'a']++;
}
for(int i = 0; i < ransomNote.size(); i++)
{
hashVec[ransomNote[i] - 'a']--;
}
bool flag = true;
for(int i = 0; i < hashVec.size(); i++)
{
if(hashVec[i] < 0)
{
flag = false;
break;
}
}
return flag;
}
};
leetcode 49 字母异位词分组
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
map<vector<int>, vector<string>> umap;
vector<vector<string>> result;
for (int i = 0; i < strs.size(); i++)
{
string str = strs[i];
vector<int> hashVec = convertStr(str);
if (umap.find(hashVec) == umap.end())
{
umap[hashVec] = vector<string>();
}
umap[hashVec].push_back(str);
}
map<vector<int>, vector<string>>::iterator it;
for (it = umap.begin(); it != umap.end(); it++)
{
result.push_back(it->second);
}
return result;
}
vector<int> convertStr(string str)
{
vector<int> hashVec(26, 0);
for (int i = 0; i < str.size(); i++)
{
hashVec[str[i] - 'a']++;
}
return hashVec;
}
};
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string, vector<string>> umap;
vector<vector<string>> result;
for (int i = 0; i < strs.size(); i++)
{
string str = strs[i];
sort(str.begin(), str.end());
if (umap.find(str) == umap.end())
{
umap[str] = vector<string>();
}
umap[str].push_back(strs[i]);
}
unordered_map<string, vector<string>>::iterator it;
for (it = umap.begin(); it != umap.end(); it++)
{
result.push_back(it->second);
}
return result;
}
};
leetcode 138 复制带随机指针的链表
class Solution {
public:
Node* copyRandomList(Node* head) {
unordered_map<Node*, int> umap; // Node address ---> vector index
vector<Node*> newNode;
Node* ptr = head;
if(!head) return NULL;
int k = 0;
while(ptr != NULL)
{
umap.insert(pair<Node*, int>(ptr, k));
newNode.push_back(new Node(ptr->val));
ptr = ptr->next;
k++;
}
ptr = head;
k = 0;
newNode.push_back(NULL); //为了在下面链接新节点的next指针不用特殊处理最后一个Node
while(ptr != NULL)
{
newNode[k]->next = newNode[k+1];
Node* tmp = ptr->random;
if(tmp != NULL)
{
int id = umap[tmp];
newNode[k]->random = newNode[id];
}
else
{
newNode[k]->random = NULL;
}
ptr = ptr->next;
k++;
}
return newNode[0];
}
};
leetcode 1002 查找共用字符
class Solution {
public:
vector<string> commonChars(vector<string>& words) {
vector<int> hashVec(26, 0);
for(int i = 0; i < words[0].size(); i++)
{
hashVec[words[0][i] - 'a']++;
}
int hashOtherStr[26] = {0};
for(int i = 1; i < words.size(); i++)
{
memset(hashOtherStr, 0, 26 * sizeof(int)); //别忘了恢复hashOtherStr的状态
for(int j = 0; j < words[i].size(); j++)
{
hashOtherStr[words[i][j] - 'a']++;
}
for(int k = 0; k < 26; k++)
{
hashVec[k] = hashVec[k] < hashOtherStr[k] ? hashVec[k] : hashOtherStr[k];
}
}
vector<string> result;
for(int k = 0; k < 26; k++)
{
while(hashVec[k])
{
string s(1, k + 'a'); //char -- string
result.push_back(s);
hashVec[k]--;
}
}
return result;
}
};
leetcode 349 两个数组的交集
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
unordered_set<int> tmpSet;
unordered_set<int> uset(nums1.begin(), nums1.end());
for(int i = 0; i < nums2.size(); i++)
{
if(uset.find(nums2[i]) != uset.end())
{
tmpSet.insert(nums2[i]);
}
}
vector<int> result(tmpSet.begin(), tmpSet.end());
return result;
}
};
leetcode 350 两个数组的交集II
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
unordered_map<int, int> umap;
for(auto ele : nums1)
{
umap[ele]++;
}
for(auto ele : nums2)
{
if(umap[ele] > 0)
{
result.push_back(ele);
umap[ele]--;
}
}
return result;
}
};
leetcode 202 快乐数
class Solution {
public:
bool isHappy(int n) {
unordered_set<int> uset;
while(1)
{
int sum = getSum(n);
if(sum == 1)
{
return true;
}
else
{
if(uset.find(sum) != uset.end())
{
return false;
}
uset.insert(sum);
n = sum;
}
}
}
int getSum(int n)
{
int sum = 0;
while(n)
{
sum += (n % 10) * (n % 10);
n = n / 10;
}
return sum;
}
};
leetcode 1 两数之和
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> mp; //key无序 key不可重复
vector<int> result;
for(int i = 0; i < nums.size(); i++)
{
auto it = mp.find(target - nums[i]); //固定一个元素，在之前的元素集合中找另一个
if(it != mp.end())
{
result.push_back(it->second);
result.push_back(i);
}
mp.insert(pair<int, int>(nums[i], i));
}
return result;
}
};
leetcode 15 三数之和
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> result;
sort(nums.begin(), nums.end());
int size = nums.size();
if (nums.empty() || nums.back() < 0 || nums.front() > 0)
{
return {};
}
for (int i = 0; i < size - 2; i++)
{
if (nums[i] > 0)
{
break;
}
if(i > 0 && nums[i] == nums[i-1])
{
continue;
}
int fix = nums[i];
int target = 0 - fix;
int first = i + 1;
int last = size - 1;
while (first < last)
{
if (nums[first] + nums[last] == target)
{
result.push_back( {nums[i], nums[first], nums[last]} );
while (first < last && nums[first] == nums[first + 1])
{
first++;
}
while (first < last && nums[last] == nums[last - 1])
{
last--;
}
first++;
last--;
}
else if (nums[first] + nums[last] < target)
{
first++;
}
else
{
last--;
}
}
}
return result;
}
};
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> result;
sort(nums.begin(), nums.end());
if(nums.size() == 0 || nums.front() > 0 || nums.back() < 0)
{
return result;
}
for(int i = 0; i < nums.size(); i++)
{
int left = i + 1;
int right = nums.size() - 1;
if(i > 0 && nums[i] == nums[i-1])
{
continue;
}
while(left < right)
{
if(nums[i] + nums[left] + nums[right] > 0)
{
right--;
}
else if (nums[i] + nums[left] + nums[right] < 0)
{
left++;
}
else
{
result.push_back(vector<int>{nums[i], nums[left], nums[right]});
while(left < right && nums[left] == nums[left+1]) // 去重逻辑应该放在找到一个三元组之后
{
left++;
}
while(left < right && nums[right] == nums[right-1])
{
right--;
}
left++; // 找到答案时，双指针同时收缩
right--;
}
}
}
return result;
}
};
leetcode 18 四数之和
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> result;
sort(nums.begin(), nums.end());
for(int i = 0; i < nums.size(); i++)
{
if(i > 0 && nums[i] == nums[i-1])
{
continue;
}
for(int k = i + 1; k < nums.size(); k++)
{
if(k > i + 1 && nums[k] == nums[k-1])
{
continue;
}
int left = k + 1;
int right = nums.size() - 1;
while(left < right)
{
//if(nums[i] + nums[k] + nums[left] + nums[right] > target) 会溢出
if(nums[i] + nums[k] > target - (nums[left] + nums[right]))
{
right--;
}
else if (nums[i] + nums[k] < target - (nums[left] + nums[right]))
{
left++;
}
else
{
result.push_back(vector<int>{nums[i], nums[k], nums[left], nums[right]});
while(left < right && nums[left] == nums[left+1]) // 去重逻辑应该放在找到一个三元组之后
{
left++;
}
while(left < right && nums[right] == nums[right-1])
{
right--;
}
left++; // 找到答案时，双指针同时收缩
right--;
}
}
}
}
return result;
}
};
leetcode 454 四数相加II
class Solution {
public:
int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
unordered_map<int, int> umap; // key --> a+b value --> a+b出现的次数
for(auto a : nums1)
{
for(auto b : nums2)
{
umap[a+b]++;
}
}
int count = 0; //计算a+b+c+d出现的次数
for(auto c : nums3)
{
for(auto d : nums4)
{
if(umap.find(0 - (c+d)) != umap.end())
{
count += umap[0 -(c+d)];
}
}
}
return count;
}
};
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