Longest Increasing Continuous subsequence I & II

Longest Increasing Continuous subsequence I 

Give you an integer array (index from 0 to n-1, where n is the size of this array),find the longest increasing continuous subsequence in this array. (The definition of the longest increasing continuous subsequence here can be from right to left or from left to right)

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

O(n) time and O(1) extra space.

Solution:

Longest Increasing Continuous subsequence II

Give you an integer matrix (with row size n, column size m),find the longest increasing continuous subsequence in this matrix. (The definition of the longest increasing continuous subsequence here can start at any row or column and go up/down/right/left any direction).

Example

Given a matrix:

<code>[
  [1 ,2 ,3 ,4 ,5],
  [16,17,24,23,6],
  [15,18,25,22,7],
  [14,19,20,21,8],
  [13,12,11,10,9]
]
</code>

return 25

Challenge

O(nm) time and memory.

Solution:

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Longest Increasing Subsequence

Longest Increasing Subsequence

Given a sequence of integers, find the longest increasing subsequence (LIS).

You code should return the length of the LIS.

Example

For [5, 4, 1, 2, 3], the LIS  is [1, 2, 3], return 3

For [4, 2, 4, 5, 3, 7], the LIS is [4, 4, 5, 7], return 4

Challenge

Time complexity O(n^2) or O(nlogn)

Clarification

What’s the definition of longest increasing subsequence?

* The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence’s elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.

* https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Solution: DP O(n^2)

lis[i]表示结尾为nums[i]的子序列的最长长度。

所以lis[i] =for all the j<i && nums[j]<=nums[i] max(lis[j])+1

其实还有可以优化为lis[i] = lis[j] +1 where nums[j] 是比nums[i]小的数里最大的一个

public int longestIncreasingSubsequence(int[] nums) {
    if (nums == null || nums.length == 0) {
        return 0;
    }
    //lis[i] means the lis of nums[0-i] which i is the last element selected in the lis
    int[] lis = new int[nums.length];
    int maxLis = 0;
    for (int i = 0; i < nums.length; i++) {
        lis[i] = 1;
        for (int j = 0; j < i; j++) {
            if (nums[i] >= nums[j]) {
                lis[i] = Math.max(lis[j] + 1, lis[i]);
            }
        }
        maxLis = Math.max(maxLis, lis[i]);
    }
    return maxLis;
}