Problem 最长连续递增／递减子序列
Give an integer array，find the longest increasing continuous subsequence in this array.
An increasing continuous subsequence:Can be from right to left or from left to right.Indices of the integers in the subsequence should be continuous.Example
For [5, 4, 2, 1, 3], the LICS is [5, 4, 2, 1], return 4.
For [5, 1, 2, 3, 4], the LICS is [1, 2, 3, 4], return 4.Note
设置正向计数器，后一位增加则计数器加1，否则置1。反向计数器亦然。
每一次比较后将较大值存入max。Solution
O(1) space, O(n) time
public class Solution { public int longestIncreasingContinuousSubsequence(int[] A) { if (A == null || A.length == 0) return 0; int n = A.length; int count = 1, countn = 1, max = 1; int i = 1; while (i != n) { if (A[i] >= A[i-1]) { count++; countn = 1; max = Math.max(max, count); } else { countn++; count = 1; max = Math.max(max, countn); } i++; } return max; }} DP using two dp arrays, O(n) space
public class Solution { public int longestIncreasingContinuousSubsequence(int[] A) { if (A == null || A.length == 0) return 0; int n = A.length; if (n == 1) return 1; int[] dp = new int[n]; int[] pd = new int[n]; int maxdp = 0, maxpd = 0; dp[0] = 1; for (int i = 1; i < n; i++) { dp[i] = A[i] >= A[i-1] ? dp[i-1]+1 : 1; maxdp = Math.max(maxdp, dp[i]); } pd[n-1] = 1; for (int i = n-2; i >= 0; i--) { pd[i] = A[i] >= A[i+1] ? pd[i+1]+1 : 1; maxpd = Math.max(maxpd, pd[i]); } return Math.max(maxdp, maxpd); }} DP using one dp arrays, O(n) space
public class Solution { public int longestIncreasingContinuousSubsequence(int[] A) { if (A == null || A.length == 0) return 0; int n = A.length; if (n == 1) return 1; int[] dp = new int[n]; int maxdp = 0, maxpd = 0; dp[0] = 1; for (int i = 1; i < n; i++) { dp[i] = A[i] >= A[i-1] ? dp[i-1]+1 : 1; maxdp = Math.max(maxdp, dp[i]); } for (int i = 1; i < n; i++) { dp[i] = A[i] <= A[i-1] ? dp[i-1]+1 : 1; maxpd = Math.max(maxpd, dp[i]); } return Math.max(maxdp, maxpd); }}