动态规划04—最长的zigzag序列

topcode链接：
https://community.topcoder.com/statc=problem_statement&pm=1259&rd=4493
problem:

A sequence of numbers is called a zig-zag sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a zig-zag sequence.For example, 1,7,4,9,2,5 is a zig-zag sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast, 1,4,7,2,5 and 1,7,4,5,5 are not zig-zag sequences, the first because its first two differences are positive and the second because its last difference is zero.Given a sequence of integers, sequence, return the length of the longest subsequence of sequence that is a zig-zag sequence. A subsequence is obtained by deleting some number of elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Examples
0)
{ 1, 7, 4, 9, 2, 5 }
Returns: 6
The entire sequence is a zig-zag sequence.
1)
{ 1, 17, 5, 10, 13, 15, 10, 5, 16, 8 }
Returns: 7
There are several subsequences that achieve this length. One is 1,17,10,13,10,16,8.
2)
{ 44 }
Returns: 1
3)
{ 1, 2, 3, 4, 5, 6, 7, 8, 9 }
Returns: 2
4)
{ 70, 55, 13, 2, 99, 2, 80, 80, 80, 80, 100, 19, 7, 5, 5, 5, 1000, 32, 32 }
Returns: 8
5)
{ 374, 40, 854, 203, 203, 156, 362, 279, 812, 955,
600, 947, 978, 46, 100, 953, 670, 862, 568, 188,
67, 669, 810, 704, 52, 861, 49, 640, 370, 908,
477, 245, 413, 109, 659, 401, 483, 308, 609, 120,
249, 22, 176, 279, 23, 22, 617, 462, 459, 244 }
Returns: 36

本题的答题思路是和LIS相同的，但是要考虑差值正负交替出现，所以我们需要一个状态去保存此时是正还是负。
（1）状态
d[i]表示数组至array[i]时，最长的zigzag长度
state[i]用来指示array[i]对比上个数是大还是小，比上一个数大则为1，小则为0,注意，此时的数不是连续的，也就是说上一个数的索引不一定是i-1，而是满足zigzag序列最长时的上一个数。

下面直接给出解决方法：

int longestZigZag(vector<int> &nums){    int N=nums.size();    vector<int> states(N,0);    vector<int> rets(N,1);    for(int i=1;i!=N;i++){        for(int j=0;j!=i;j++){            if((states[j]==0)||(states[j]==-1&&nums[i]>nums[j])||(states[j]==1&&nums[i]<nums[j]))                if(rets[j]+1>rets[i]){                    rets[i]=rets[j]+1;                    if(nums[i]>nums[j])                        states[i]=1;                    if(nums[i]<nums[j])                        states[i]=-1;                }        //std::cout<<i<<": "<<rets[i]<<"states: "<<states[i]<<std::endl;        }    }    return rets[N-1];}