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Preface
- FAQ
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Part I - Basics
- Basics Data Structure
- Basics Sorting
- Basics Algorithm
- Basics Misc
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Part II - Coding
- String
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Integer Array
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Remove Element
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Zero Sum Subarray
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Subarray Sum K
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Subarray Sum Closest
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Recover Rotated Sorted Array
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Product of Array Exclude Itself
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Partition Array
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First Missing Positive
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2 Sum
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3 Sum
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3 Sum Closest
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Remove Duplicates from Sorted Array
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Remove Duplicates from Sorted Array II
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Merge Sorted Array
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Merge Sorted Array II
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Median
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Partition Array by Odd and Even
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Kth Largest Element
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Remove Element
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Binary Search
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First Position of Target
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Search Insert Position
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Search for a Range
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First Bad Version
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Search a 2D Matrix
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Search a 2D Matrix II
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Find Peak Element
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Search in Rotated Sorted Array
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Search in Rotated Sorted Array II
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Find Minimum in Rotated Sorted Array
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Find Minimum in Rotated Sorted Array II
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Median of two Sorted Arrays
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Sqrt x
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Wood Cut
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First Position of Target
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Math and Bit Manipulation
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Single Number
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Single Number II
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Single Number III
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O1 Check Power of 2
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Convert Integer A to Integer B
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Factorial Trailing Zeroes
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Unique Binary Search Trees
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Update Bits
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Fast Power
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Hash Function
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Happy Number
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Count 1 in Binary
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Fibonacci
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A plus B Problem
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Print Numbers by Recursion
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Majority Number
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Majority Number II
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Majority Number III
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Digit Counts
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Ugly Number
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Plus One
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Palindrome Number
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Task Scheduler
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Single Number
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Linked List
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Remove Duplicates from Sorted List
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Remove Duplicates from Sorted List II
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Remove Duplicates from Unsorted List
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Partition List
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Add Two Numbers
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Two Lists Sum Advanced
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Remove Nth Node From End of List
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Linked List Cycle
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Linked List Cycle II
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Reverse Linked List
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Reverse Linked List II
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Merge Two Sorted Lists
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Merge k Sorted Lists
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Reorder List
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Copy List with Random Pointer
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Sort List
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Insertion Sort List
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Palindrome Linked List
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LRU Cache
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Rotate List
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Swap Nodes in Pairs
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Remove Linked List Elements
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Remove Duplicates from Sorted List
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Binary Tree
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Binary Tree Preorder Traversal
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Binary Tree Inorder Traversal
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Binary Tree Postorder Traversal
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Binary Tree Level Order Traversal
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Binary Tree Level Order Traversal II
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Maximum Depth of Binary Tree
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Balanced Binary Tree
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Binary Tree Maximum Path Sum
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Lowest Common Ancestor
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Invert Binary Tree
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Diameter of a Binary Tree
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Construct Binary Tree from Preorder and Inorder Traversal
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Construct Binary Tree from Inorder and Postorder Traversal
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Subtree
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Binary Tree Zigzag Level Order Traversal
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Binary Tree Serialization
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Binary Tree Preorder Traversal
- Binary Search Tree
- Exhaustive Search
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Dynamic Programming
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Triangle
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Backpack
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Backpack II
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Minimum Path Sum
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Unique Paths
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Unique Paths II
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Climbing Stairs
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Jump Game
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Word Break
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Longest Increasing Subsequence
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Palindrome Partitioning II
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Longest Common Subsequence
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Edit Distance
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Jump Game II
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Best Time to Buy and Sell Stock
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Best Time to Buy and Sell Stock II
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Best Time to Buy and Sell Stock III
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Best Time to Buy and Sell Stock IV
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Distinct Subsequences
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Interleaving String
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Maximum Subarray
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Maximum Subarray II
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Longest Increasing Continuous subsequence
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Longest Increasing Continuous subsequence II
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Maximal Square
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Triangle
- Graph
- Data Structure
- Big Data
- Problem Misc
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Part III - Contest
- Google APAC
- Microsoft
- Appendix I Interview and Resume
-
Tags
Longest Words
Question
- lintcode: (133) Longest Words
Given a dictionary, find all of the longest words in the dictionary.
Example
Given
{
"dog",
"google",
"facebook",
"internationalization",
"blabla"
}
the longest words are(is) ["internationalization"].
Given
{
"like",
"love",
"hate",
"yes"
}
the longest words are ["like", "love", "hate"].
Challenge
It's easy to solve it in two passes, can you do it in one pass?
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题解
简单题,容易想到的是首先遍历以便,找到最长的字符串,第二次遍历时取最长的放到最终结果中。但是如果只能进行一次遍历呢?一次遍历意味着需要维护当前遍历的最长字符串,这必然有比较与更新删除操作,这种情况下使用双端队列最为合适,这道题稍微特殊一点,不必从尾端插入,只需在遍历时若发现比数组中最长的元素还长时删除整个列表。
Java
class Solution {
/**
* @param dictionary: an array of strings
* @return: an arraylist of strings
*/
ArrayList<String> longestWords(String[] dictionary) {
ArrayList<String> result = new ArrayList<String>();
if (dictionary == null || dictionary.length == 0) return result;
for (String str : dictionary) {
// combine empty and shorter length
if (result.isEmpty() || str.length() > result.get(0).length()) {
result.clear();
result.add(str);
} else if (str.length() == result.get(0).length()) {
result.add(str);
}
}
return result;
}
}
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源码分析
熟悉变长数组的常用操作。
复杂度分析
时间复杂度 , 最坏情况下需要保存 n - 1个字符串,空间复杂度 .