Sword finger offer special assault version day 3

The finger of the sword Offer II 007. Array and 0 Three numbers of

Ideas

Enumerate the first number $i$ , And then let $target$ by $-nums[i]$, Finally, find the sum of all two numbers through double pointers $target$ The combination of . Time complexity $O(n^2)$ .
doubt ： How to ensure that there are no duplicate combinations ？
Combine the idea of de duplication with 47. Full Permutation II equally
practice ： Sort in ascending order first , If the second number and the third number are equal to the previous number respectively and are not being used , Then the current number cannot be used , If used, there will be repeated sequences . Instead, you can use .

class Solution {
    
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
    
        int n = nums.size();
        vector<vector<int>> res;
        sort(nums.begin(),nums.end());
        for(int i = 0; i < n-2; i++) {
    
            if(i!=0 && nums[i] == nums[i-1]) continue;
            int target = -nums[i];
            // Double pointer finds and is target All non repeated combinations of the two numbers of 
            for(int j = i+1, k = n-1; j < k; j++) {
    
                if(nums[j] == nums[j-1] && j-1 != i) continue;
                if(nums[j] + nums[k] > target) 
					k--, j--;
                else if(nums[j] + nums[k] == target)
                	res.push_back({
    nums[i],nums[j],nums[k]});
            }
        }
        return res;
    }
};

The finger of the sword Offer II 008. And greater than or equal to target The shortest subarray of

Ideas

Seeing and seeking Continuous subarray The length of , intuition -> Double pointer , Pay attention to both ends of the subarray .

class Solution {
    
public:
    int minSubArrayLen(int target, vector<int>& nums) {
    
        int n = nums.size();
        int i = 0, j = 0, res = 0x3f3f3f3f;
        int sum = 0;
        while(i <= j && j < n){
    
            if(sum + nums[j] < target) sum += nums[j++];
            else res = min(res,j-i+1), sum -= nums[i++];
        }
        if(res == 0x3f3f3f3f) return 0;
        else return res;
    }
};

The finger of the sword Offer II 009. The product is less than K Subarray

Ideas

See what the title requires Number of consecutive subarrays , Intuition should be done with double pointers .
Sure enough , Just pay attention to both ends of the continuous subarray .

class Solution {
    
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
    
        int n = nums.size();
		int i = 0, j = 0, cnt = 0;
        int product = 1;
        while(i <= j && j < n){
    
            if(nums[j] * product < k) {
    
                cnt += j-i+1; // If you have any questions, you can count them by hand 
                product *= nums[j++];
            } else {
    
                product /= nums[i++];
            }
        }
        return cnt;
    }
};