package LeetCode_300 /** * 300. Longest Increasing Subsequence * https://leetcode.com/problems/longest-increasing-subsequence/description/ * Given an unsorted array of integers, find the length of longest increasing subsequence. Example: Input: [10,9,2,5,3,7,101,18] Output: 4 Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. Note: 1. There may be more than one LIS combination, it is only necessary for you to return the length. 2. Your algorithm should run in O(n2) complexity. Follow up: Could you improve it to O(n log n) time complexity? * */ class Solution2 { /* * solution 2: Patience Sorting, Time:O(nlogn), Space:O(n) * https://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/LongestIncreasingSubsequence.pdf * */ fun lengthOfLIS(nums: IntArray): Int { val n = nums.size if (n == 0) { return 0 } val list = ArrayList<Int>() for (i in nums.indices) { /* //method 1: Collections.binarySearch: * 1. list contains nums[i], return the index of nums[i] in list; * 2. list not contains nums[i], return ~index, the index is the first number larger than nums[i], * (if key not found, index start from 1); * */ /*var x = Collections.binarySearch(list, nums[i]) if (x < 0) { x = x.inv() } if (x == list.size) { list.add(nums[i]) } else { list.set(x, nums[i]) }*/ //method 2: if (list.isEmpty() || list.get(list.lastIndex) < nums[i]) { list.add(nums[i]) } else { list.set(findFirstLargeOrEquals(list, nums[i]), nums[i]) } } return list.size } private fun findFirstLargeOrEquals(list: ArrayList<Int>, target: Int): Int { var left = 0 var right = list.size - 1 while (left < right) { val mid = left + (right - left) / 2 if (list.get(mid) < target) { //search in right side left = mid + 1 } else { right = mid } } return left } }