DAY7打卡|454.四数相加II 、 383. 赎金信 、15. 三数之和 、18. 四数之和

454.四数相加II 

复杂度感觉还是很高

class Solution {
public:
    int fourSumCount(vector& nums1, vector& nums2, vector& nums3, vector& nums4) 
    {
        unordered_mapmp;
        for(int a:nums1)
        {
            for(int b:nums2)
            mp[a+b]++;
        }
        int count=0;
        for(int c:nums3)
        {
            for(int d:nums4)
            {
                if(mp.find(0-(c+d))!=mp.end())
                count+=mp[0-(c+d)];
            }
        }
        return count;
    }
};

  383. 赎金信 

数据量比较小的情况下用数组处理比较节约时间。

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) 
    {
       // unordered_mapmp;
       int record[26]={0};
       // for(int i=0;imagazine.size())
       return false;
       for(int i=0;i

 15. 三数之和 

class Solution {
public:
    vector> threeSum(vector& nums) 
    {
        vector>ans;
        if(nums.size()<3)
        return ans;
        sort(nums.begin(),nums.end());
        for(int k=0;k0)
            return ans;
          if( k>0&&nums[k]==nums[k-1])
          continue;

            int i=k+1,j=nums.size()-1;
          
            while(i0)
            j--;
            else if(s<0)
            i++;
            else
            {
                ans.push_back({nums[k],nums[i],nums[j]});
                while(j>i&&nums[j]==nums[j-1])
                j--;
                while(j>i&&nums[i]==nums[i+1])
                i++;
                j--;
                i++;


            }

            }

        }
return ans ;
    }
};

// 这道题目使用双指针要比哈希表高效
class Solution {
public:
    vector> threeSum(vector& nums) {
        vector> ans;
        if (nums.size() < 3) return ans;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0) return ans;
            // 去重
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int left = i + 1;
            int right = nums.size() - 1;
            while (left < right) {
                if (nums[left] + nums[right] + nums[i] > 0) {
                    right--;
                } else if (nums[left] + nums[right] + nums[i] < 0){
                    left++;
                } else {
                    ans.push_back({nums[left], nums[right], nums[i]});
                    while (right > left && nums[right] == nums[right - 1]) right--;
                    while (right > left && nums[left] == nums[left + 1]) left++;
                    left++;
                    right--;
                }
            }
        }
        return ans;
    }
};

 18. 四数之和 

class Solution {
public:
    vector> fourSum(vector& nums, int target) 
    {
        sort(nums.begin(),nums.end());
        vector>res;
        if(nums.size()<4)
        return res;
        int a,b,c,d,n=nums.size();
        for(a=0;a<=n-4;a++)
        {
            if(a>0&&nums[a]==nums[a-1])
            continue;
            for(b=a+1;b<=n-3;b++)
            {
                if(b>a+1&&nums[b]==nums[b-1])
                continue;
                c=b+1,d=n-1;
                while(ccur_target)
                    d--;
                    else
                    {
                        res.push_back({nums[a],nums[b],nums[c],nums[d]});
                        while(c