返回最长上升子序列

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(
4000
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;

}
}