返回最长上升子序列

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返回最长子序列
题目要求：
给定一个整数数组（下标从 0 到 n-1， n 表示整个数组的规模），请找出该数组中的最长上升连续子序列。（最长上升连续子序列可以定义为从右到左或从左到右的序列。）

样例
给定 [5, 4, 2, 1, 3], 其最长上升连续子序列（LICS）为 [5, 4, 2, 1], 返回 4.

给定 [5, 1, 2, 3, 4], 其最长上升连续子序列（LICS）为 [1, 2, 3, 4], 返回 4.

version1:
step1:
先把输入数组当中的后一位数减去前一位数的结果存在一个vector(deviation)中，这样我们就可以通过判断数的正负，从而判断大小序列关系了；
step2:
统计deviation当中连续的序列，最后返回最大值即可；

int LICS(vector<int> & A){    if (A.empty())    {        return 0;    }    vector<int> deviation;    for (int i = 1; i < A.size(); i++)    {        deviation.push_back(A[i] - A[i - 1]);    }    int count_positive = 0;    vector<int> positive;    vector<int> negtive;    int count_negtive = 0;    for (int i = 0; i < deviation.size(); i++)    {        if (deviation[i] < 0)        {            if (count_positive != 0)            {                positive.push_back(count_positive);            }            count_positive = 0;            count_negtive++;        }        if (count_negtive != 0)        {            negtive.push_back(count_negtive);        }        if (deviation[i] > 0)        {            if (count_negtive != 0)            {                negtive.push_back(count_negtive);            }            count_negtive = 0;            count_positive++;        }        if (count_positive != 0)        {            positive.push_back(count_positive);        }    }    vector<int>::iterator biggest_positive = max_element(positive.begin(), positive.end());    if (biggest_positive != positive.end())    {        count_positive = ((*biggest_positive) + 1);    }    vector<int>::iterator biggest_negative = max_element(negtive.begin(), negtive.end());    if (biggest_negative != negtive.end())    {        count_negtive = ((*biggest_negative) + 1);    }    if (count_positive > count_negtive)    {        return count_positive;    }    return count_negtive;}

version2:
通过正向和反向的连续遍历，统计最大上升序列；

public class Solution {    /**     * @param A an array of Integer     * @return  an integer     */    public int longestIncreasingContinuousSubsequence(int[] A) {           int max = 1,count = 1;           if(A.length <=0){               return A.length;           }           //正向遍历           for(int i = 1; i<A.length; ){               while(i<A.length && A[i]>A[i-1] ){                   i++;                   count++;               }if(count > max){                   max = count;               }               count = 1;               i++;           }           //反向遍历           for(int i = A.length - 1 ; i >= 0; ){               while(i > 0 && A[i]<A[i-1]  ){                   i--;                   count++;               }if(count > max){                   max = count;               }               count = 1;               i--;           }           return max;       }}

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