4.1 MaxProductOfThree

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);        }    }