哈希表02

454. 四数相加 II

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        int result = 0;
        unordered_map<int, int> mp;
        for(int i = 0; i < nums1.size(); i++)
        {
            for(int j = 0; j < nums2.size(); j++)
            {
               
                    mp[nums1[i] + nums2[j]]++;
            }
        }
        for(int i = 0; i < nums3.size(); i ++)
        {
            for(int j = 0; j < nums4.size(); j++)
            {
                if(mp.find(-nums3[i] - nums4[j]) != mp.end())
                    result += mp[-nums3[i] - nums4[j]];
            }
        }
        return result;
    }
};

383. 赎金信

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int record[26] = {0};
        //add
        if (ransomNote.size() > magazine.size()) {
            return false;
        }
        for (int i = 0; i < magazine.length(); i++) {
            // 通过record数据记录 magazine里各个字符出现次数
            record[magazine[i]-'a'] ++;
        }
        for (int j = 0; j < ransomNote.length(); j++) {
            // 遍历ransomNote，在record里对应的字符个数做--操作
            record[ransomNote[j]-'a']--;
            // 如果小于零说明ransomNote里出现的字符，magazine没有
            if(record[ransomNote[j]-'a'] < 0) {
                return false;
            }
        }
        return true;
    }
};

15. 三数之和

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> results;
        sort(nums.begin(), nums.end());
        for(int i = 0; i < nums.size(); i++)
        {
            if(nums[i] > 0) return results;
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int left = i + 1;
            int right = nums.size() - 1;
            while(left < right)
            {
                if(nums[i] + nums[left] + nums[right] < 0)
                    left++;
                else if(nums[i] + nums[left] + nums[right] > 0)
                    right--;
                else
                {
                    results.push_back(vector<int>{nums[i], nums[left], nums[right]});
                    while(left < right && nums[left] == nums[left + 1]) left++;
                    while(left < right && nums[right] == nums[right - 1]) right--;
                    left++;
                    right--;
                }
            }
        }
        return results;

    }
};

18. 四数之和

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> results;
        sort(nums.begin(), nums.end());
        for(int i = 0; i < nums.size(); i++)
        {
            if(nums[i] > target && nums[i] >= 0) return results;
            if(i > 0 && nums[i] == nums[i - 1]) continue;
            for(int j = i + 1; j < nums.size(); j++)
            {
                 if (nums[j] + nums[i] > target && nums[j] + nums[i] >= 0) {
                    break;
                }

                // 对nums[i]去重
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                int left = j + 1; int right = nums.size() - 1;
                while(left < right)
                {
                    if((long)nums[i] + nums[j] + nums[left] + nums[right] < target)
                        left++;
                    else if((long)nums[i] + nums[j] + nums[left] + nums[right] > target)
                        right--;
                    else
                    {
                        results.push_back(vector<int>{nums[i] , nums[j] , nums[left] , nums[right]});
                        while(left < right && nums[right - 1] == nums[right]) right--;
                        while(left < right && nums[left + 1] == nums[left]) left++;
                        right--; left++;
                    }
                }
            }
        }
        return results;
    }
};