Summary of most consistency problems

Catalog

1. Sum of two numbers

167. Sum of two numbers II - Input an ordered array

653. Sum of two numbers IV - Input BST

560. And for K Subarray

523. Continuous subarrays and

713. The product is less than K Subarray

152. Product maximum subarray

15. Sum of three numbers

16. The closest sum of three

18. Sum of four numbers

454. Add four numbers II

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1. Sum of two numbers

Given an array of integers nums  And an integer target value target, Please find... In the array And is the target value target   the   Two   Integers , And return their array subscripts .

You can assume that each input corresponds to only one answer . however , The same element in the array cannot be repeated in the answer .

You can return the answers in any order .

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        std::unordered_map <int,int> map;
        for(int i = 0; i < nums.size(); i++) {
            auto iter = map.find(target - nums[i]);
            if(iter != map.end()) {
                return {iter->second, i};
            }
            map.insert(pair<int, int>(nums[i], i));
        }
        return {};
    }
};

167. Sum of two numbers II - Input an ordered array

I'll give you a subscript from 1 The starting array of integers  numbers , The array has been pressed Non decreasing order   , Please find out from the array that the sum satisfying the addition is equal to the target number  target Two numbers of . Let these two numbers be numbers[index1] and numbers[index2] , be 1 <= index1 < index2 <= numbers.length .

In length 2 Array of integers for [index1, index2] Returns the subscript of these two integers in the form of index1 and index2.

You can assume that each input Only corresponding to the only answer , And you Can not be Reuse the same elements .

The solution you design must use only constant level extra space .

class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
    vector<int> res;
    int l=0,r=numbers.size()-1;
    while(true){
        int sum=numbers[l]+numbers[r];
        if(sum==target) {
            res.push_back(l+1);
            res.push_back(r+1);
            break;
        }
        else if(sum>target){
            r--;
        }
        else l++;
    }

    return res;
    }
};

653. Sum of two numbers IV - Input BST

Given a binary search tree  root  And a target result  k, If BST There are two elements in and their sum is equal to the given target result , Then return to  true.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    unordered_set<int> set;
    bool findTarget(TreeNode* root, int k) {
        if(root==nullptr) return false;
        if(set.count(k-root->val)) return true;
        else set.insert(root->val);
        return findTarget(root->left,k)||findTarget(root->right,k);
    }
};

560. And for K Subarray

Give you an array of integers  nums  And an integer  k , Please count and return   The array is  k  The number of consecutive subarrays of  .

The prefix and

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        mp[0] = 1;
        int count = 0, pre = 0;
        for(int i=0;i<nums.size();i++){
            pre+=nums[i];
            if(mp.count(pre-k)){
                count+=mp[pre-k];
            }
            mp[pre]++;
        }
        return count;
    }
};

523. Continuous subarrays and

（ Not fully mastered ）

Give you an array of integers nums And an integer  k , Write a function to determine whether the array contains continuous subarrays that meet the following conditions at the same time ：

Subarray size At least for 2 , And
The sum of subarray elements is k Multiple .
If there is , return true ; otherwise , return false .

If there is an integer n , Make integer x accord with x = n * k , said x yes k A multiple of .0 Always regarded as k A multiple of .

【 Congruence theorem 】 【 Hashtable 】【 Simplify prefixes and 】

Congruence theorem ： If two integers m、n Satisfy n-m Can be k to be divisible by , that n and m Yes k congruence

namely ( pre(j) - pre (i) ) % k == 0 be pre(j) % k == pre(i) % k

deduction => pre (i) % k = (a0 + a1 + ... + ai) % k = (a0 % k + a1 % k + ... ai % k ) % k （ This derivation is useful in simplifying prefixes and , State the current prefix and % k It will not affect the following prefixes and % k ）

Hashtable Storage Key ：pre(i) % k
Value： i

Ergodic process ：

1. Calculate the prefix and pre( j ) % k

2. When pre(j) % k Already exists in the hash table , It means that there is i Satisfy pre(j) % k == pre(i) % k ( i < j )

HashMap in , It is known that Key, You can get it. Value namely i Value , Last Judge j - i >= 2 Is it true that will do

     3. When pre(j) % k Does not exist in the hash table , Will (pre(j) % k, j ) Save to hash table         

class Solution {
public:
    unordered_map<int,int> map;
    bool checkSubarraySum(vector<int>& nums, int k) {
        if(nums.size()<2) return false;
        for(int i=1;i<nums.size();i++){
            nums[i]+=nums[i-1];// The prefix and 
        }
        map[0]=-1;// initialization   Prefixes and are 0 The subscript of is -1  non-existent 
        for(int i=0;i<nums.size();i++){
            int temp=nums[i]%k;
            // if(temp==0) return true;
            if(map.count(temp)) {
                int preindex=map[temp];
                if(i-preindex>=2) return true;
            }
            else map[temp]=i;
        }
        return false;
    }
};

713. The product is less than K Subarray

Give you an array of integers  nums  And an integer  k , Please return that the product of all elements in the subarray is strictly less than  k  The number of consecutive subarrays of .

Example 1：

Input ：nums = [10,5,2,6], k = 100
Output ：8
explain ：8 The product is less than 100 The subarrays of are ：[10]、[5]、[2],、[6]、[10,5]、[5,2]、[2,6]、[5,2,6].
It should be noted that [10,5,2] It's not that the product is less than 100 Subarray .

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        int n = nums.size(), ret = 0;
        int prod = 1;
        int left=0;
        for(int right=0;right<nums.size();right++){// The right pointer traverses from left to right 
            prod*=nums[right];// Calculate the sum of the products in the sliding window 
            while(left<=right&&prod>=k){
                prod/=nums[left++];// The sum of prefix products is greater than or equal to k  Move the left pointer to the right   And remove the current element of the left pointer 
            }
            ret+=right-left+1;// The number of subarrays in the sliding window  
        }
        return ret;
    }
};

152. Product maximum subarray

Give you an array of integers nums , Please find the non empty continuous subarray with the largest product in the array （ The subarray contains at least one number ）, And returns the product of the subarray .

The answer to the test case is  32- position Integers .

Subarray Is a continuous subsequence of an array .

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int maxF = nums[0], minF = nums[0], ans = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            // Maximum and minimum values before storage 
            int mx=maxF;
            int mn=minF;

            maxF=max(maxF*nums[i],max(nums[i],mn*nums[i]));// Similar to dynamic planning 
            minF=min(minF*nums[i],min(nums[i],mx*nums[i]));

            ans = max(maxF, ans);
        }
        return ans;
    }
};

15. Sum of three numbers

Give you one containing n Array of integers  nums, Judge  nums  Are there three elements in a,b,c , bring  a + b + c = 0 ？ Please find out all and for 0 Triple without repetition .

Be careful ： Answer cannot contain duplicate triples .

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> result;
        for(int i=0;i<nums.size();i++){
            if (i > 0 && nums[i] == nums[i-1])  continue;// duplicate removal   Otherwise, there will be repeated solutions 
            int left=i+1;
            int right=nums.size()-1;
            while(left<right){
                if(nums[left]+nums[right]>-nums[i]) right--;
                else if(nums[left]+nums[right]<-nums[i]) left++;
                else{
                    result.push_back({nums[i],nums[left],nums[right]});
                    left++;
                    right--;
                    // After finding the solution   same left and right No more use   Need to be heavy  while loop 
                    while(left<right&&nums[left]==nums[left-1]) left++;
                    while(left<right&&nums[right]==nums[right+1]) right--;
                }
            }
        }
        return result;
    }
};

16. The closest sum of three

Give you a length of n Array of integers for  nums  and A target value  target. Please start from nums Choose three integers , Make their sum with  target  Nearest .

Returns the sum of the three numbers .

It is assumed that there is only exactly one solution for each set of inputs .

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        int best = 1e8;
        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){
                int sum=nums[left]+nums[right]+nums[i];
                if(sum==target) return sum;
                if(abs(sum-target)<abs(best-target)){
                    best=sum;
                }
                if(sum>target){
                    right--;
                    while(left<right&&nums[right]==nums[right+1]) right--;
                }
                if(sum<target){
                    left++;
                    while(left<right&&nums[left]==nums[left-1]) left++;
                }

            }

        }

        return best;
    }
};

18. Sum of four numbers

Here you are n An array of integers  nums , And a target value target . Please find and return the quads that meet all the following conditions and are not repeated  [nums[a], nums[b], nums[c], nums[d]] （ If two Quad elements correspond to each other , It is considered that two quads repeat ）：

0 <= a, b, c, d < n
a、b、c and d Different from each other
nums[a] + nums[b] + nums[c] + nums[d] == target
You can press In any order Return to the answer .

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 j=i+1;j<nums.size();j++){
                if(j>i+1&&nums[j]==nums[j-1]) continue;
                int left=j+1;
                int right=nums.size()-1;
                while(left<right){// There has to be this while loop 
                    long int temp=nums[i]+nums[j];
                    if((long)nums[right]+nums[left]>target-temp){
                        right--;
                    }
                    else if((long)nums[right]+nums[left]<target-temp){
                        left++;
                    }
                    else {
                        result.push_back({nums[i],nums[j],nums[left],nums[right]});
                        right--;
                        left++;
                        while(left<right&&nums[left]==nums[left-1]) left++;
                        while(left<right&&nums[right]==nums[right+1]) right--;
                    }
                }
            }
        }
        return result;
    }
};

454. Add four numbers II

Here are four integer arrays nums1、nums2、nums3 and nums4 , The length of the array is n , Please calculate how many tuples there are (i, j, k, l) To meet the ：

0 <= i, j, k, l < n
nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int, int> umap; //key:a+b The numerical ,value:a+b The number of times the value appears 
        //  Ergodic large A And big B Array , Count the sum of two array elements , And the number of times , Put it in map in 
        for (int a : A) {
            for (int b : B) {
                umap[a + b]++;
            }
        }
        int count = 0; //  Statistics a+b+c+d = 0  Number of occurrences 
        //  In ergodic big C And big D Array , Find out if  0-(c+d)  stay map If you've seen it in the past , Just put map in key Corresponding value That is, the number of occurrences .
        for (int c : C) {
            for (int d : D) {
                if (umap.find(0 - (c + d)) != umap.end()) {
                    count += umap[0 - (c + d)];
                }
            }
        }
        return count;
    }
};