Leetcode find three numbers and satisfy the conditions xx topic summary 15➕16➕259

 

The most basic is the subject of a two-pointer range where to find two numbers is equal to Target. First, the interval from small to large. Next, le long as a pointer, a pointer ri, respectively, from the left border to the middle section to advance. Complexity is the sort of nlogn.

Here are a few questions to find three numbers in a range of conditions and satisfy xx. The subject of the first practice or the first order. After a first digit is fixed, then the right section of the figure before re-use to find the two numbers and the algorithm. I.e., i from 0 to n-1, j is increased from +. 1 i, k from n-1 starts to decrease until the collision j and k. Such a complexity of n squared. Layer i from 0 to n-1 is O (n), from the inner layer i + 1 to n-1 is also O (n).

15

 1 class Solution {
 2 public:
 3     vector<vector<int>> threeSum(vector<int>& nums) {
 4         if(nums.empty()){return vector<vector<int>>();}
 5         sort(nums.begin(),nums.end());
 6         vector<vector<int>> res;
 7         int n=nums.size();
 8         for(int i=0;i<n;++i){
 9             if(i>0 and nums[i-1]==nums[i]){continue;}
10             int j=i+1,k=n-1;
11             while(j<k){
12                 while(j<k and j>i+1 and nums[j]==nums[j-1]){
13                     ++j;
14                 }
15                 while(j<k and k<n-1 and nums[k]==nums[k+1]){
16                     --k;
17                 }
18                 if(j>=k){break;}
19                 if(nums[i]+nums[j]+nums[k]>0){
20                     --k;
21                 }
22                 else if(nums[i]+nums[j]+nums[k]<0){
23                     ++j;
24                 }
25                 else{
26                     res.push_back({nums[i],nums[j],nums[k]});
27                     ++j,--k;
28                 }
29             }
30         }
31         return res;
32     }
33 };

 

16

 1 class Solution {
 2 public:
 3     int threeSumClosest(vector<int>& nums, int target) {
 4         int n=nums.size();
 5         if(n<3){return 0;}
 6         sort(nums.begin(),nums.end());
 7         int min_distance=INT_MAX,temp,sum=0;
 8         for(int i=0;i<n;++i){
 9             if(i>0 and nums[i]==nums[i-1]){continue;}
10             int j=i+1,k=n-1;
11             while(j<k){
12                 if((temp=nums[i]+nums[j]+nums[k]-target)>0){
13                     if(temp<min_distance){
14                         min_distance=temp;
15                         sum=temp+target;
16                     }
17                     --k;
18                     while(j<k and nums[k]==nums[k+1]){
19                         --k;
20                     }
21                 }
22                 else if(temp<0){
23                     if(-temp<min_distance){
24                         min_distance=-temp;
25                         sum=temp+target;
26                     }
27                     ++j;
28                     while(j<k and nums[j-1]==nums[j]){
29                         ++j;
30                     }
31                 }
32                 else{
33                     return target;
34                 }
35             }
36         }
37         return sum;
38     }
39 };

 

259

 1 class Solution {
 2 public:
 3     int threeSumSmaller(vector<int>& nums, int target) {
 4         int N=nums.size(),res=0;
 5         sort(nums.begin(),nums.end());
 6         for(int i=0;i<N-2;++i){
 7             int j=i+1,k=N-1;
 8             while(j<k){
 9                 if(nums[i]+nums[j]+nums[k]<target){
10                     res+=k-j;
11                     j++;
12                 }
13                 else{
14                     k--;
15                 }
16             }
17         }
18         return res;
19     }
20 };