Longest Increasing Subsequence

Description:

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

For "ABCD" and "EACB", the LCS is "AC", return 2.

Challenge

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

Clarification

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

Solution:

class Solution {
public:
	/**
	 * @param nums: The integer array
	 * @return: The length of LIS (longest increasing subsequence)
	 */
	int longestIncreasingSubsequence(vector<int> nums) {
		auto sz = (int)nums.size();
		if (sz == 0) return 0;
		vector<int> dp(sz, 0);
		vector<int> cost(sz+1, 0); // const[i] = min{nums[t], dp[t] = i}
		dp[0] = 1;
		cost[1] = nums[0];
		int cnt = 1; // counter of cost
		int rc = 1;
		for (int i = 1; i < sz; ++i) {
			if (nums[i] < cost[1]) {
				cost[1] = nums[i];
				dp[i] = 1;
			} else if (nums[i] >= cost[cnt]) {
				++cnt;
				cost[cnt] = nums[i];
				dp[i] = cnt;
			} else {
				int k = binarySearch(cost, 1, cnt, nums[i]);
				cost[k] = nums[i];
				dp[i] = k;
			}
			rc = max(rc, dp[i]);
		}
		return rc;
	}

private:
	int binarySearch(vector<int>& nums, int l, int r, int target) {
		auto sz = (int)nums.size();
		if (sz == 0) return 0;
		while (l <= r) {
			int mid = l+r>>1;
			if (nums[mid] > target) r = mid-1;
			else if (nums[mid] < target) l = mid+1;
			else return mid;
		}
		return l;
	}
};