leetcode-hash

两数之和

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> record;
        for(int i = 0 ; i < nums.size() ; i ++){
            int complement = target - nums[i];
            if(record.find(complement) != record.end()){
                int res[] = {i, record[complement]};
                return vector<int>(res, res + 2);
            }
            record[nums[i]] = i;
        }
    }

无重复字符的最长子串

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int freq[256] = {0};
        int l = 0, r = -1; //滑动窗口为s[l...r]
        int res = 0;
        // 整个循环从 l == 0; r == -1 这个空窗口开始
        // 到l == s.size(); r == s.size()-1 这个空窗口截止
        // 在每次循环里逐渐改变窗口, 维护freq, 并记录当前窗口中是否找到了一个新的最优值
        while(l < s.size()){
            if(r + 1 < s.size() && freq[s[r+1]] == 0){
                r++;
                freq[s[r]]++;
            }else {   //r已经到头 || freq[s[r+1]] == 1
                freq[s[l]]--;
                l++;
            }
            res = max(res, r-l+1);
        }
        return res;
    }
};


class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int res = 0, left = -1, n = s.size();
        unordered_map<int, int> m;
        for (int i = 0; i < n; ++i) {
            if (m.count(s[i]) && m[s[i]] > left) {
                left = m[s[i]];  
            }
            m[s[i]] = i;
            res = max(res, i - left);            
        }
        return res;
    }
};

三数之和

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        if (nums.empty() || nums.back() < 0 || nums.front() > 0) return {};
        for (int k = 0; k < nums.size(); ++k) {
            if (nums[k] > 0) break;
            if (k > 0 && nums[k] == nums[k - 1]) continue;
            int target = 0 - nums[k];
            int i = k + 1, j = nums.size() - 1;
            while (i < j) {
                if (nums[i] + nums[j] == target) {
                    res.push_back({nums[k], nums[i], nums[j]});
                    while (i < j && nums[i] == nums[i + 1]) ++i;
                    while (i < j && nums[j] == nums[j - 1]) --j;
                    ++i; --j;
                } else if (nums[i] + nums[j] < target) ++i;
                else --j;
            }
        }
        return res;
    }
};

重复的dna序列

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        vector<string> res;
        if (s.size() <= 10) return res;
        int mask = 0x7ffffff, cur = 0;
        unordered_map<int, int> m;
        for (int i = 0; i < 9; ++i) {
            cur = (cur << 3) | (s[i] & 7);
        }
        for (int i = 9; i < s.size(); ++i) {
            cur = ((cur & mask) << 3) | (s[i] & 7);
            if (m.count(cur)) {
                if (m[cur] == 1) res.push_back(s.substr(i - 9, 10));
                ++m[cur]; 
            } else {
                m[cur] = 1;
            }
        }
        return res;
    }
};

两个数组的交集

// 时间复杂度: O(nlogn)
// 空间复杂度: O(n)
class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        set<int> record;
        for( int i = 0 ; i < nums1.size() ; i ++ ){
            record.insert(nums1[i]);
        }
        set<int> resultSet;
        for( int i = 0 ; i < nums2.size() ; i ++ ){
            if(record.find(nums2[i]) != record.end()){
                resultSet.insert(nums2[i]);
            }
        }
        vector<int> resultVector;
        for(set<int>::iterator iter = resultSet.begin(); iter != resultSet.end(); iter ++ ){
            resultVector.push_back(*iter);
        }

        return resultVector;
    }
};

两个数组的交集 II

// 时间复杂度: O(nlogn)
// 空间复杂度: O(n)
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        map<int, int> record;
        for(int i = 0 ; i < nums1.size() ; i ++){
             record[nums1[i]] += 1;
        }
        vector<int> resultVector;
        for(int i = 0 ; i < nums2.size() ; i ++){
            if(record[nums2[i]] > 0){
                resultVector.push_back(nums2[i]);
                record[nums2[i]] --;
            }
        }
        return resultVector;
    }
};

回旋镖的数量

// 时间复杂度: O(n^2)
// 空间复杂度: O(n)
class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int res = 0;
        for( int i = 0 ; i < points.size() ; i ++ ){
            // record中存储 点i 到所有其他点的距离出现的频次
            unordered_map<int, int> record;
            for(int j = 0 ; j < points.size() ; j ++){
                if(j != i){
                    // 计算距离时不进行开根运算, 以保证精度
                    record[dis(points[i], points[j])] += 1;
                }
            }
            for(unordered_map<int, int>::iterator iter = record.begin() ; iter != record.end() ; iter ++){
                res += (iter->second) * (iter->second - 1);
            }
        }
        return res;
    }

private:
    int dis(const pair<int,int> &pa, const pair<int,int> &pb){
        return (pa.first - pb.first) * (pa.first - pb.first) +
               (pa.second - pb.second) * (pa.second - pb.second);
    }
};

四个数相加

// 时间复杂度: O(n^2)
// 空间复杂度: O(n^2)
class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int,int> hashtable;
        for(int i = 0 ; i < A.size() ; i ++){
            for(int j = 0 ; j < B.size() ; j ++){
                 hashtable[A[i]+B[j]] += 1;
            }
        }
        int res = 0;
        for(int i = 0 ; i < C.size() ; i ++){
            for(int j = 0 ; j < D.size() ; j ++){
                if(hashtable.find(-C[i]-D[j]) != hashtable.end()){
                    res += hashtable[-C[i]-D[j]];
                }
            }
        }
        return res;
    }
};