Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
问题描述: Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else…
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, ksuch that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
1. Description Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤…
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
Description: Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-…
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
Question Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 el…
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array. Formally the function should: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return…
这个题是说看一个没有排序的数组里面有没有三个递增的子序列,也即: Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false. 大家都知道这个题有很多解法,然而题主丧心病狂地说要O(n)的时间复杂度和O(1)的空间复杂度. 我当时考虑的是找三个递增的数,中间那个数比较重要,所以我们可以遍历该数组,检查每个元素是不是…
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. Notice O(n…
[题目描述] 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. Not…
http://www.lintcode.com/en/problem/longest-increasing-continuous-subsequence/# 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…
DFS + Memorized Search (DP) class Solution { int dfs(int i, int j, int row, int col, vector<vector<int>>& A, vector<vector<int>>& dp) { ) return dp[i][j]; && A[i-][j] > A[i][j]) { dp[i][j] = max(dp[i][j], dfs(i -…
mycode time limited class Solution(object): def increasingTriplet(self, nums): """ :type nums: List[int] :rtype: bool """ length = len(nums) temp = [] for i,num_i in enumerate(nums[:length-2]): for j,num_j in enumerate(nums…
309. Best Time to Buy and Sell Stock with Cooldown class Solution { public int maxProfit(int[] prices) { if(prices == null || prices.length <= 1) return 0; int n = prices.length; int[] hold = new int[n]; int[] unhold = new int[n]; hold[0] = -prices[0…
169. Majority Element Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Credit…
LinkedIn(39) 1 Two Sum 23.0% Easy 21 Merge Two Sorted Lists 35.4% Easy 23 Merge k Sorted Lists 23.3% Hard 33 Search in Rotated Sorted Array 30.2% Hard 34 Search for a Range 29.1% Medium 46 Permutations 35.7% Medium 47 Permutations II 28.0% Medium 50…
刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…
突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…