Find three numbers in N numbers to minimize the absolute value of their sum-

Original question:

Write a function to find any subset of 3 integers from a set of N integers that have a sum with the smallest absolute value,And analyze the time and space complexity of your function.

The input to this function is a pointer or reference to an array of integers, along with the value of N.
When N> = 3, the output is a 3-element array of integers, such that their sum has the smallest absolute value.
When n <3, the output is null.

For example, given the following input:

{-99,-66, 0, 2, 3}, your function shocould return {0, 2, 3 }.

 

Meaning:

Given the unordered array s Containing N numbers (may contain negative numbers, 0, positive numbers ). Calculate the numbers A, B, and C to minimize the absolute values of sum.
Example: S = {-9, 0, 6}, A =-9, B = 3, C = 6, min = 0

 

 

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Method 1,The brute-force search method, which contains three numbers in triplicate loops and the time complexity is O (n ^ 3). The Code is as follows:

 

 // The time complexity is O (n ^ 3) <br/> // x, y, in Z, three selected numbers are stored in sequence as the subscript in the original array <br/> void brutaldeal (int A [], int N) <br/>{< br/> int x, y, z; <br/> int min = 1000000; <br/> for (INT I = 0; I <N; I ++) <br/>{< br/> for (Int J = I + 1; j <n; j ++) <br/> for (int K = J + 1; k <n; k ++) <br/> If (ABS (A [I] + A [J] + A [k]) <min) <br/>{< br/> X = I; <br/> Y = J; <br/> Z = K; <br/> min = ABS (A [I] + A [J] + A [k]); <br/>}< br/> cout <Endl <"min = A [" <x <"] + A [" <Y <"] + A [" <z <"] =" <A [x] <"+" <A [y] <"+" <<A [Z] <"=" <min <Endl; <br/>}

 

Method 2,After sorting array A, enumerate two of them and search for another number in two parts. O (N ^ 2 * logn)

 

 

Method 3,After sorting array A, enumerate the number of cur (A [I]), and use the two-way pointer head and tail to update from both ends of the array (remove cur ). Update the head or tail based on the positive and negative numbers of a [head] + A [tail] + cur. If it is negative, head ++; otherwise, tail --. Once the sum is 0, exit. At the same time, Min is updated based on the sum. The last min is the request. Use three variables X, Y, and Z to record the subscript of any optimal solution in the sorted array. The worst time complexity of this method is O (n ^ 2 ). First write a fast sorting algorithm, so the code is as follows:

 

// Sort the elements from a [s] to a [T] Quickly <br/> // call format: quicksort (A, 0, n-1). A is an array, n is the array length <br/> void quicksort (int A [], int S, int t) <br/>{< br/> int I = s, j = T; <br/> int TMP; <br/> If (S <t) <br/> {<br/> TMP = A [s]; // TMP temporary Start Element <br/> while (I! = J) <br/>{< br/> while (j> I & A [J]> = TMP) // find the first number smaller than TMP from right to left <br/> j --; <br/> A [I] = A [J]; // move a [J] to the position of a [I] <br/> while (I <J & A [I] <= TMP) // find the first number larger than TMP from left to right <br/> I ++; <br/> A [J] = A [I]; // move a [I] to the position of a [J] <br/>}< br/> A [I] = TMP; // move the starting element TMP to the location of a [I] <br/> quicksort (A, S, I-1); // I is the subscript of the split point, A [I] is a partition element <br/> quicksort (A, I + 1, t ); <br/>}</P> <p> void sortdeal (int A [], int N) <br/>{< br/> // sort the array first <br/> quicksort (A, 0, n-1); <br/> for (INT I = 0; I <n; I ++) <br/> cout <A [I] <""; </P> <p> // start selection <br/> int min = 1000000; <br/> int cur; <br/> int head, tail; <br/> int x, y, z; <br/> for (INT I = 0; I <n; I ++) <br/>{< br/> head = 0; <br/> tail = n-1; </P> <p> cur = A [I]; <br/> while (Head <tail) <br/> {<br/> If (Head = I) <br/> head ++; // remove cur (A [I]) <br/> If (tail = I) <br/> tail --; // remove cur (A [I]) <br/> int now = cur + A [head] + A [tail]; <br/> If (now = 0 | ABS (now) <min) <br/>{< br/> min = ABS (now); <br/> X = I; <br/> Y = head; <br/> Z = tail; <br/> If (now = 0) <br/> break; <br/>}< br/> else if (now <0) <br/> head ++; <br/> else <br/> tail --; <br/>}< br/> If (min = 0) <br/> break; <br/>}< br/> cout <Endl <"min = A [" <x <"] + A [" <Y <"] + A ["<z <"] = "<A [x] <" + "<A [y] <" + "<A [Z] <"=" <min <Endl; <br/>}< br/>

 

 

 

 

 

The main functions of the preceding method are as follows:

 

 

# Include <iostream> <br/> # include <time. h> <br/> using namespace STD; </P> <p> int main () <br/> {<br/> int N; <br/> cout <"input N:"; <br/> CIN> N; <br/> int * ap = new int [N]; <br/> srand (unsigned) Time (null )); </P> <p> // randomly generate elements in the array AP [N] <br/> for (INT I = 0; I <n; I ++) <br/>{< br/> aP [I] = rand (); <br/> If (RAND () % 2) <br/> aP [I] * =-1; <br/> cout <AP [I] <""; <br/>}</P> <p> brutaldeal (AP, n); <br/> sortdeal (AP, n); </P> <p> return 0; <br/>}