目录
1、1. 两数之和
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> hash;
for (int i = 0; i < nums.size(); i++) {
int x = target - nums[i];
if (hash.count(x))
return {hash[x], i};
hash[nums[i]] = i;
}
return {-1, -1};
}
};2、面试题 01.02. 判定是否互为字符重排
class Solution {
public:
bool CheckPermutation(string s1, string s2) {
if (s1.size() != s2.size())
return false;
int hash[26] = {0};
for (auto s : s1)
hash[s - 'a']++;
for (auto s : s2) {
hash[s - 'a']--;
if (hash[s - 'a'] < 0)
return false;
}
return true;
}
};3、217. 存在重复元素
class Solution {
public:
bool containsDuplicate(vector<int>& nums) {
unordered_set<int> hash;
for (auto n : nums) {
if (hash.count(n))
return true;
else
hash.insert(n);
}
return false;
}
};4、 219. 存在重复元素 II
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
unordered_map<int, int> hash;
for (int i = 0; i < nums.size(); i++) {
if (hash.count(nums[i])) {
if (i - hash[nums[i]] <= k)
return true;
}
hash[nums[i]] = i;
}
return false;
}
};5、49. 字母异位词分组
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string, vector<string>> hash;
for (auto& s : strs) {
string tmp = s;
sort(tmp.begin(), tmp.end());
hash[tmp].push_back(s);
}
vector<vector<string>> ret;
for (auto& x : hash) {
ret.push_back(x.second);
}
return ret;
}
};
