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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
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Tags
Bitmap
最开始接触 bitmap 是在《编程珠玑》这本书上,书中所述的方法有点简单粗暴,不过思想倒是挺好——从信息论的角度来解释就是信息压缩了。即将原来32位表示一个 int 变为一位表示一个 int. 从空间的角度来说就是巨大的节省了(1/32)。可能的应用有大数据排序/查找(非负整数)。核心思想为根据最大非负整数确定位数,对应的位依次排序。
C++ 中有bitset
容器,其他语言可用类似方法实现。
Implementation
C
#include <stdio.h>
#include <stdlib.h>
/*
* @param bits: uint array, i: num i of original array
*/
void setbit(unsigned int *bits, unsigned int i, int BIT_LEN)
{
bits[i / BIT_LEN] |= 1 << (i % BIT_LEN);
}
/*
* @param bits: uint array, i: num i of original array
*/
int testbit(unsigned int *bits, unsigned int i, int BIT_LEN)
{
return bits[i / BIT_LEN] & (1 << (i % BIT_LEN));
}
int main(int argc, char *argv[])
{
const int BIT_LEN = sizeof(int) * 8;
const unsigned int N = 1 << (BIT_LEN - 1);
unsigned int *bits = (unsigned int *)calloc(N / BIT_LEN, sizeof(int));
for (unsigned int i = 0; i < N; i++) {
if (i % 10000001 == 0) setbit(bits, i, BIT_LEN);
}
for (unsigned int i = 0; i < N; i++) {
if (testbit(bits, i, BIT_LEN) != 0) printf("i = %u exists.\n", i);
}
free(bits);
bits = NULL;
return 0;
}
copy
源码分析
核心为两个函数方法的使用,setbit
用于将非负整数i
置于指定的位。可用分区分位的方式来理解位图排序的思想,即将非负整数i
放到它应该在的位置。比如16,其可以位于第一个 int 型的第17位,具体实现即将第17位置一,细节见上面代码。测试某个数是否存在于位图中也可以采用类似方法。