Leetcode (3Sum Closest)

Topic

Given an array S of n integers, find three integers in S such, the sum was closest to a Given number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Analysis

This question is very similar to the previous one. The sum of the ternary elements closest to the target value in an integer array sequence is obtained.
The solution to this problem is a three-layer traversal, a summation comparison, but it is obvious that the time-out exception will inevitably occur when Leetcode commits. So you have to find another method, the AC code of this topic comes from the reference, here to express our thanks.

Time Limit exceeded code

classSolution { Public://Method 1:3-level exhaustive, timeout exception    intThreesumclosest ( vector<int>& Nums,intTarget) {intLen = Nums.size ();if(Len <3)        {return 0; }intclosest = nums[0] + nums[1] + nums[2]; for(inti =1; I < len-2; i++) {//First go to the re-operation            if(Nums[i] = = Nums[i-1])Continue; for(intj = i +1; J < Len-1; J + +) {if(Nums[j] = = Nums[j-1])Continue; for(intK = j +1; K < Len; k++) {if(Nums[k] = = Nums[k-1])Continue;intsum = Nums[i] + nums[j] + nums[k];if(sum = = target) {closest = sum; Break; }//if                    Else    if(ABS(Sum-target) <ABS(Closest-target))                    {closest = sum; }//elif                    Else                        Continue; }//for}//for}//for        returnClosest }int ABS(intA) {returnA >0? A:0-A; }};

AC Code

 class solution { Public:intThreesumclosest (vector<int>& Nums,intTarget) {size_t size = nums.size ();if(Size <3) {cout <<"num size must bigger than there!"<< Endl;return 0; } sort (Nums.begin (), Nums.end ());//For the following process must be sorted in advance, similar to binary search        intresult =0;//Record the final result        intDistance = numeric_limits<int>::max ();//Signed int        int sum=0;//Intermediate resultssize_t i =0, j = i +1, k = Size-1; for(i =0; I < size-2; i++)//The first element of the ternary group is traversed one time, with a range of [0...n-3]{//de-duplication to avoid repeated calculations, if and last the same skip            if(I >0&& Nums[i] = = Nums[i-1])            {Continue; } j = i +1;//After the first element of the ternary group is selected, the second element is examined from the next position of the first elementK = Size-1;//After the first element of the ternary group is selected, the third element is examined from the end of the array             while(J < K)The last two elements of the ternary group use the left and right approximation to skip efficiency, and after the first element is selected, all subsequent elements are examined only once.{sum= Nums[i] + nums[j] + nums[k];if(sum= = target)///existence distance of the nearest 0 is returned directly, otherwise the exhaustive selection is not 0 minimum distance{return sum; }Else if(sum< target) {if(Target-sum) (< distance) {result =sum; Distance = target-sum; } j = j +1;//Avoid repeated calculations if you skipped with the last time                    if(Nums[j] = = Nums[j-1]) {j = j +1; }                }Else if(sum> Target) {if((sum-target) < distance) {result =sum; Distance =sum-Target; } k = k-1;//Avoid repeated calculations if you skipped with the last time                    if(Nums[k] = = Nums[k +1]) {k = k-1; }                }            }        }returnResult }};

GitHub test Program source code

Leetcode (3Sum Closest)