Maximize A[P] * A[Q] * A[R] for any triplet (P, Q, R).
寻找数组A,三个数乘积的最大值。
A non-empty zero-indexed array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R] (0 ≤ P < Q < R < N).
For example, array A such that:
A[0] = -3
A[1] = 1
A[2] = 2
A[3] = -2
A[4] = 5
A[5] = 6
contains the following example triplets:
(0, 1, 2), product is −3 * 1 * 2 = −6
(1, 2, 4), product is 1 * 2 * 5 = 10
(2, 4, 5), product is 2 * 5 * 6 = 60
Your goal is to find the maximal product of any triplet.
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty zero-indexed array A, returns the value of the maximal product of any triplet.
For example, given array A such that:
A[0] = -3
A[1] = 1
A[2] = 2
A[3] = -2
A[4] = 5
A[5] = 6
the function should return 60, as the product of triplet (2, 4, 5) is maximal.
Assume that:
N is an integer within the range [3..100,000];
each element of array A is an integer within the range [−1,000..1,000].
Complexity:
expected worst-case time complexity is O(N*log(N));
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Solution
import java.util.*;
class Solution {
public int solution(int[] A) {
Arrays.sort(A);
int end = A.length-1;
return Math.max(A[0]*A[1]*A[end], A[end-2]*A[end-1]*A[end]);
}
}
主要考虑负负得正,除此之外均是最后三个数乘积最大。这里我偷了个懒吼吼,用了java的sort肯定满足nlogn。下面补上快排
A[p,q]数组分为三个区A[p,lower]<=key, A[high,q]>key,(lower,high)待排序
public int partition(int[] A, int p, int q){
int lower=p;
int high=q;
int key = A[(p+q)/2];
while(lower<=high){
while(A[lower]<key){
lower++;
}
while(A[high]>key){
high++;
}
if(lower<=high){
swap(A,lower++,high--);
}
}
return lower;//lower=high, A[low]=key
}
public void quicksort(int[] A, int left, int right){
if(left<right){
int index = partition(A,left,right);
quicksort(A,left,index-1);
quicksort(A,index+1,right);
}
}