8.3 Peaks

A non-empty zero-indexed array A consisting of N integers is given.

A peak is an array element which is larger than its neighbors. More precisely, it is an index P such that 0 < P < N − 1, A[P − 1] < A[P] and A[P] > A[P + 1].

For example, the following array A:

A[0] = 1A[1] = 2A[2] = 3A[3] = 4A[4] = 3A[5] = 4A[6] = 1A[7] = 2A[8] = 3A[9] = 4A[10] = 6A[11] = 2

has exactly three peaks: 3, 5, 10.

We want to divide this array into blocks containing the same number of elements. More precisely, we want to choose a number K that will yield the following blocks:

A[0], A[1], …, A[K − 1],
A[K], A[K + 1], …, A[2K − 1],
…
A[N − K], A[N − K + 1], …, A[N − 1].
What’s more, every block should contain at least one peak. Notice that extreme elements of the blocks (for example A[K − 1] or A[K]) can also be peaks, but only if they have both neighbors (including one in an adjacent blocks).

The goal is to find the maximum number of blocks into which the array A can be divided.

Array A can be divided into blocks as follows:

one block (1, 2, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2). This block contains three peaks.
two blocks (1, 2, 3, 4, 3, 4) and (1, 2, 3, 4, 6, 2). Every block has a peak.
three blocks (1, 2, 3, 4), (3, 4, 1, 2), (3, 4, 6, 2). Every block has a peak. Notice in particular that the first block (1, 2, 3, 4) has a peak at A[3], because A[2] < A[3] > A[4], even though A[4] is in the adjacent block.
However, array A cannot be divided into four blocks, (1, 2, 3), (4, 3, 4), (1, 2, 3) and (4, 6, 2), because the (1, 2, 3) blocks do not contain a peak. Notice in particular that the (4, 3, 4) block contains two peaks: A[3] and A[5].

The maximum number of blocks that array A can be divided into is three.

Write a function:

class Solution { public int solution(int[] A); }

that, given a non-empty zero-indexed array A consisting of N integers, returns the maximum number of blocks into which A can be divided.

If A cannot be divided into some number of blocks, the function should return 0.

For example, given:

A[0] = 1A[1] = 2A[2] = 3A[3] = 4A[4] = 3A[5] = 4A[6] = 1A[7] = 2A[8] = 3A[9] = 4A[10] = 6A[11] = 2

the function should return 3, as explained above.

Assume that:

N is an integer within the range [1..100,000];
each element of array A is an integer within the range [0..1,000,000,000].
Complexity:

expected worst-case time complexity is O(N*log(log(N)));
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
将数组拆分为等长子数组，保证每个至少包含一个peak。首先找出所有peak保存起来。子数组长度必须能被数组长度整除，然后按照从大到小的顺序穷举，直到找到一个长度满足条件。因为peak的index是有序的，所以内层循环可以采用二分查找。
注意java Arrays的二分查找的结果包括头，但是不包括尾。所以我在peaks的末尾加入一个辅助数组。
虽然这个算法是满分，但是我觉得还不够简洁，希望有更好的方案

class Solution {    public int solution(int[] A) {       if(A.length<3) return 0;       int[] peaks = new int[A.length];       int length = -1;       for(int i=1; i<A.length-1;){            if(A[i-1]<A[i] && A[i]>A[i+1]){                peaks[++length] = i;                i+=2;            }            else{                i++;            }        }        peaks[++length]=A.length;        for(int i=3; i<=A.length; i++){            if(A.length%i == 0){                int rank = i;                int block =  A.length/rank;                if(length < block/2                   || peaks[0]>rank-1){                    continue;                }                 else{                    int j;                    for(j=1; j<block;j++){                       int start = Arrays.binarySearch(peaks,0,length,rank*j);        //System.out.println("i:" + i + "j:"+ j +"start:"+start);                       if(start>=0){                            continue;                                               }                        else{                            int end = Arrays.binarySearch(peaks,0,length,rank*(j+1)-1);                               // System.out.println("i:" + i + "j:"+ j +"end:"+end);                            if(start == end){                                break;                            }                        }                    }                    if(j==block) return block;                }            }        }        return 0;    }}