Wiggle Subsequence

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problem description:

A sequence of numbers is called a wiggle 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 wiggle sequence.

For example, [1,7,4,9,2,5] is a wiggle 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 wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.

Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

解决方案：

假设前i个数已经找到了他们的最大摇摆子序列，若下一个数该选择一个小于前面的数，则应该选择第i个数后面中从大到小排列中最小的数；若下一个数该选择大于前面的数，则选择第i个数后面有小到大排列子序列中最大的数。

第一个数应该选择连续三个有序中中间的数。

显然这种方法对每个数判断一次，复杂度为O（n）。

不过用leetcode提交显示当输入序列为[33,53,12,64,50,41,45,21,97,35,47,92,39,0,93,55,40,46,69,42,6,95,51,68,72,9,32,84,34,64,6,2,26,98,3,43,30,60,3,68,82,9,97,19,27,98,99,4,30,96,37,9,78,43,64,4,65,30,84,90,87,64,18,50,60,1,40,32,48,50,76,100,57,29,63,53,46,57,93,98,42,80,82,9,41,55,69,84,82,79,30,79,18,97,67,23,52,38,74,15]时，正确结果应该是67，我的结果是63.还没找到问题所在。

在网上找了找别人的方法后才发现自己的方法真复杂，我太蠢了，把问题搞得真复杂。

代码如下：

class Solution {public:    int wiggleMaxLength(vector<int>& nums) {vector<int>::iterator it1=nums.begin();vector<int>::iterator it2=nums.end();int i,len,d1,d2,d3,d4,result;int temp;len=it2-it1;if(len<=1)//序列中只有一个数或0个数return len;else if(len==2){if(*it1==*(it1+1))//输入两个相同的数值return 1;else//输入两个不同的值return 2;}else{i=1;while(*it1==*(it1+i)&&i<len){i++;}if(i==len){if(*it1==*(it1+i-1))return 1;elsereturn 2;}else{/*d1=*(it1+i)-*it1;result=2;it1=it1+i;temp=*it1;it1++;*//*d2=*(it1+i)-*it1;d1=-d2;temp=*it1;it1=it1+i;result=1;*/bool m,n;while((it1+2)!=it2){m=(*it1<=*(it1+1))&&(*(it1+1)<=*(it1+2));n=(*it1>=*(it1+1))&&(*(it1+1)>=*(it1+2));//if((*it1<=*(it1+1)<=*(it1+2))||(*it1>=*(it1+1)>=*(it1+2)))if(m||n){//temp=*(it1+1);it1=it1+1;}else break;}d2=*(it1+1)-*it1;d1=-d2;temp=*it1;it1=it1+1;result=1;while(it1!=it2){d2=*it1-temp;if(d1*d2<0){result++;if((it1+1)!=it2){d3=*(it1+1)-temp;if(d1*d3<0){d4=*(it1+1)-*it1;if(abs(d3)>abs(d2)&&(d4*d2>0)){it1++;temp=*it1;d1=d3;}else{temp=*it1;    it1++;    d1=d2;}}else{    temp=*it1;    d1=d2;it1++;//break;}}elseit1++;}elseit1++;}return result;}}    }};

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