题目：
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.
解题：
这个直接找就可以了，最长的升序，和最长的降序，再求最大值，时间复杂度O(N)
Java程序：
public class Solution {
/**
* @param A an array of Integer
* @return an integer
*/
public int longestIncreasingContinuousSubsequence(int[] A) {
// Write your code here
int left = Integer.MIN_VALUE;
int right = Integer.MIN_VALUE;
int m = A.length;
int tmp1 = 1;
int tmp2 = 1;
if (m==0)
return 0;
for(int i = 0; i<m-1 ;i++){
if (A[i] <= A[i+1]){
tmp1++;
}else{
left = left>tmp1?left:tmp1;
tmp1 = 1;
}
if (A[i] >= A[i+1]){
tmp2 ++;
}else{
right = right>tmp2?right:tmp2;
tmp2 = 1;
}
}
left = left>tmp1?left:tmp1;
right = right>tmp2?right:tmp2;
return left>right?left:right;
}
}
总耗时: 6578 ms
Python程序：
class Solution:
# @param {int[]} A an array of Integer
# @return {int} an integer
def longestIncreasingContinuousSubsequence(self, A):
# Write your code here
m = len(A)
if m ==0:
return 0
left = 0
right = 0
tmp1 = 1
tmp2 = 1
for i in range(m-1):
if A[i]<= A[i+1]:
tmp1+=1
else:
left = max(left,tmp1)
tmp1 = 1
if A[i]>= A[i+1]:
tmp2+=1
else:
right = max(right,tmp2)
tmp2 = 1
left = max(tmp1,left)
right = max(tmp2,right)
return max(left, right)
总耗时: 546 ms