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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,5 @@ | ||
| { | ||
| "title": "The Art of Programming By July", | ||
| "description": "本书是July和他伙伴们的《程序员编程艺术》的电子书", | ||
| "coverimage": "TAOP.png" | ||
| } |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,120 @@ | ||
| /** | ||
| * Copyright (c) 2014 The TAOPP book Authors. All rights reserved. | ||
| * Use of this source code is governed by a BSD-style license that can be found | ||
| * in the LICENSE file. | ||
| * | ||
| * Filename: LCAProblem.c | ||
| * Brief: The C program for solving LCA problem | ||
| * Version: 0.1 | ||
| * Created: Tue Mar 18 21:25:58 2014 | ||
| * | ||
| * Author(s): Liang Wang<fairywell28@yahoo.com> | ||
| * eriol | ||
| * July | ||
| * | ||
| */ | ||
|
|
||
| /** | ||
| * @brief 在二叉排序树中查找给定两个节点的LCA节点 | ||
| * | ||
| * @author C version 0.1, 2014/3/18, Liang Wang<fairywell28@yahoo.com> | ||
| * C++ version 0.2, 2014, July | ||
| * C++ version 0.1, 2011, eriol | ||
| * @brief 在二叉排序树中查找给定两个节点的LCA节点 | ||
| * | ||
| * @param root 该二叉排序树的根节点指针 | ||
| * @param u 给定节点一的指针 | ||
| * @param v 给定节点二的指针 | ||
| * | ||
| * @return 找到LCA节点返回其指针；否则返回NULL | ||
| */ | ||
| Node* FindLowestCommonAncestorBst(Node* root, Node* u, Node* v) | ||
| { | ||
| int left_value = u->value; | ||
| int right_value = v->value; | ||
| Node* parent_node = NULL; | ||
| Node* cur_node = root; | ||
|
|
||
| // 参数检查 | ||
| if (NULL == root || NULL == u || NULL == v | ||
| || root == u || root == v || u == v) { | ||
| fprintf(stderr, "Wrong input data! Exit!"); | ||
| return NULL; | ||
| } | ||
|
|
||
| // 调整左右节点值到正确 | ||
| if (left_value > right_value) { | ||
| swap(left_value, right_value); | ||
| } | ||
|
|
||
| while (cur_node) { | ||
| // 如果cur_node的值小于u、v的值，说明LCA节点应该在其右子树中 | ||
| if (cur_node->value < left_value) { | ||
| parent_node = cur_node; | ||
| cur_node = cur_node->right; | ||
| } else if (cur_node->value > right_value) { | ||
| // 如果cur_node的值大于u、v的值，则应该查询其左子树 | ||
| parent_node = cur_node; | ||
| cur_node = cur_node->left; | ||
| } else if (cur_node->value == left_value || cur_node->value == right_value) { | ||
| // 找到节点u或者v处，说明其父节点即为所求 | ||
| return parent_node; | ||
| } else { | ||
| // 现在cur_node的值处于u和v的值之间，显然即为所求 | ||
| return cur_node; | ||
| } | ||
| } | ||
|
|
||
| // 数据有误，找不到任何的LCA节点 | ||
| return NULL; | ||
| } | ||
|
|
||
| /** | ||
| * @brief 在二叉排序树中查找给定两个节点的LCA节点，递归版本 | ||
| * | ||
| * @author C version 0.1, 2014/3/19, Liang Wang<fairywell28@yahoo.com> | ||
| * | ||
| * @param root 该二叉排序树的根节点指针 | ||
| * @param u 给定节点一的指针 | ||
| * @param v 给定节点二的指针 | ||
| * | ||
| * @return 找到LCA节点返回其指针；否则返回NULL | ||
| */ | ||
| Node* FindLcaBstRecursively(Node* root, Node* u, Node* v) | ||
| { | ||
| // 参数检查 | ||
| // NOTICE:特别包含了u、v即是root节点的情况，此时应该返回NULL | ||
| if (NULL == root || NULL == u || NULL == v | ||
| || root == u || root == v || u == v) { | ||
| fprintf(stderr, "Wrong input data: Arguments check failed! Exit!"); | ||
| return NULL; | ||
| } | ||
|
|
||
| int left_value = u->value; | ||
| int right_value = v->value; | ||
|
|
||
| // 调整左右节点值到正确 | ||
| if (left_value > right_value) { | ||
| swap(left_value, right_value); | ||
| } | ||
|
|
||
| // 判断当前节点是否为所求LCA节点 | ||
| // 情形1：u、v分别在root的左右子树上 | ||
| if (root->value > left_value | ||
| && root->value < right_value) { | ||
| return root; | ||
| } | ||
|
|
||
| // 情形2：u、v都在root的同一棵子树且u、v有一个节点是root的子节点 | ||
| if (root->right == u || root->right == v | ||
| || root->left == u || root->left == v) { | ||
| return root; | ||
| } | ||
|
|
||
| // 当前节点不是所求LCA，则递归返回左右子树的LCA节点 | ||
| if (root->value < left_value) { // 查找右子树 | ||
| return FindLcaBstRecursively(root->right, u, v); | ||
| } else if (root->value > right_value) { // 查找左子树 | ||
| return FindLcaBstRecursively(root->left, u, v); | ||
| } | ||
| } |
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| //compiled with g++ | ||
| #include <iostream> | ||
| #include <vector> | ||
| #include <algorithm> | ||
| #include <iterator> | ||
| #include <cassert> | ||
| using namespace std; | ||
|
|
||
| void cycle_leader(vector<int> &a,int from, int mod) { | ||
| for (int i = from * 2 % mod;i != from; i = i * 2 % mod) | ||
| swap(a[from],a[i]); | ||
| } | ||
|
|
||
| void perfect_shuffle(vector<int> &a,int n){ | ||
| int n2,m,i,k,t; | ||
| vector<int>::iterator iter = a.begin()+1; //exclude index 0 | ||
| for(;n > 1;){ | ||
| //step 1 | ||
| n2 = n * 2; | ||
| for(k = 0,m = 1; n2 / m >=3; ++k,m *= 3) | ||
| ; | ||
| m /= 2; | ||
| // 2m = 3^k - 1 , 3^k <= 2n < 3^(k + 1) | ||
|
|
||
| //step 2 STL/algorithm-rotate | ||
| rotate(iter+m, iter+n, iter+m+n ); //right cyclic shift of the index[m+1,...,n+m] O(n) | ||
|
|
||
| //step 3 | ||
| for(i = 0,t = 1;i < k; ++i,t *= 3) | ||
| cycle_leader(a,t,m * 2 +1); | ||
|
|
||
| //step 4 | ||
| iter += (m * 2); | ||
| n -= m; | ||
| } | ||
| //n = 1 | ||
| swap(a[1],a[2]); | ||
| } | ||
|
|
||
| int main(){ | ||
| vector<int > a; | ||
| a.push_back(0);//to make index start from 1 | ||
| int num = 0,length = 0; | ||
| cout<<"Please input your number to be shuffled and '0' to end up \n"; | ||
| cin>>num; | ||
| while(num != 0){ | ||
| a.push_back(num); | ||
| cin>>num; | ||
| } | ||
| cout<<"The input number is \n"; | ||
| copy(a.begin()+1,a.end(),ostream_iterator<int> (cout," ")); | ||
| cout<<"\n"; | ||
|
|
||
| length = a.size(); | ||
| assert(length % 2 == 1); //the size of input number is even | ||
| perfect_shuffle(a,length / 2); | ||
|
|
||
| cout<<"After shuffered,the number is \n"; | ||
| copy(a.begin()+1,a.end(),ostream_iterator<int> (cout," ")); | ||
| cout<<"\n"; | ||
| } |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,157 @@ | ||
| /* | ||
| * By Frederick-S | ||
| * 2014-03-08 | ||
| */ | ||
|
|
||
| // 1. 穷举 | ||
|
|
||
| /* | ||
| * @param {Array} arr 目标数组 | ||
| * @param {Number} number 在数组中查找两个数，使得其和为 number | ||
| * @example | ||
| * var arr = solution1([1, 2], 3); | ||
| */ | ||
| function solution1(arr, number) { | ||
| for (var i = 0, length = arr.length; i < length; i++) { | ||
| for (var j = i + 1; j < length; j++) { | ||
| if (arr[i] + arr[j] == number) { | ||
| return [arr[i], arr[j]]; | ||
| } | ||
| } | ||
| } | ||
|
|
||
| return []; | ||
| } | ||
|
|
||
| // 2. 二分查找 | ||
|
|
||
| /* | ||
| * @param {Array} arr 目标数组 | ||
| * @param {Number} number 在数组中查找两个数，使得其和为 number | ||
| * @example | ||
| * var arr = solution2([1, 2], 3); | ||
| */ | ||
| function solution2(arr, number) { | ||
| arr.sort(function (a, b) { | ||
| return a - b; | ||
| }); | ||
|
|
||
| var index = 0; | ||
| for (var i = 0, length = arr.length; i < length; i++) { | ||
| index = binarySearch(arr, number - arr[i], 0, length - 1); | ||
|
|
||
| if (index >= 0 && index != i) { | ||
| return [arr[i], arr[index]]; | ||
| } | ||
| } | ||
|
|
||
| return []; | ||
| } | ||
|
|
||
| function binarySearch(arr, key, low, high) { | ||
| var mid = 0; | ||
|
|
||
| while (low <= high) { | ||
| mid = Math.floor((low + high) / 2); | ||
|
|
||
| if (arr[mid] == key) { | ||
| return mid; | ||
| } else if (arr[mid] > key) { | ||
| high -= 1; | ||
| } else if (arr[mid] < key) { | ||
| low += 1; | ||
| } | ||
| } | ||
|
|
||
| return -1; | ||
| } | ||
|
|
||
| // 3. 双数组指针扫描 | ||
|
|
||
| /* | ||
| * @param {Array} arr 目标数组 | ||
| * @param {Number} number 在数组中查找两个数，使得其和为 number | ||
| * @example | ||
| * var arr = solution3([1, 2], 3); | ||
| */ | ||
| function solution3(arr, number) { | ||
| var subtraction = arr.map(function (value, index, array) { | ||
| return number - value; | ||
| }); | ||
|
|
||
| arr.sort(function (a ,b) { | ||
| return a - b; | ||
| }) | ||
|
|
||
| subtraction.sort(function (a, b) { | ||
| return b - a; | ||
| }) | ||
|
|
||
| var i = 0, length = arr.length, j = length - 1; | ||
|
|
||
| while (i < length && j > -1) { | ||
| if (arr[i] < subtraction[j]) { | ||
| i++; | ||
| } else if (arr[i] > subtraction[j]) { | ||
| j--; | ||
| } else { | ||
| if (i != j) { | ||
| return [arr[i], number - arr[i]]; | ||
| } | ||
| } | ||
| } | ||
|
|
||
| return []; | ||
| } | ||
|
|
||
| // 4. 借助 hash | ||
|
|
||
| /* | ||
| * @param {Array} arr 目标数组 | ||
| * @param {Number} number 在数组中查找两个数，使得其和为 number | ||
| * @example | ||
| * var arr = solution4([1, 2], 3); | ||
| */ | ||
| function solution4(arr, number) { | ||
| var hash = {}; | ||
|
|
||
| arr.forEach(function (value, index, array) { | ||
| hash[value] = value; | ||
| }); | ||
|
|
||
| for (var i = 0, length = arr.length; i < length; i++) { | ||
| if (hash[number - arr[i]]) { | ||
| return [arr[i], number - arr[i]]; | ||
| } | ||
| } | ||
|
|
||
| return []; | ||
| } | ||
|
|
||
| // 5. 双指针两端扫描 | ||
|
|
||
| /* | ||
| * @param {Array} arr 目标数组 | ||
| * @param {Number} number 在数组中查找两个数，使得其和为 number | ||
| * @example | ||
| * var arr = solution5([1, 2], 3); | ||
| */ | ||
| function solution5(arr, number) { | ||
| var i = 0, j = arr.length - 1; | ||
|
|
||
| arr.sort(function (a, b) { | ||
| return a - b; | ||
| }) | ||
|
|
||
| while (i < j) { | ||
| if (arr[i] + arr[j] > number) { | ||
| j--; | ||
| } else if (arr[i] + arr[j] < number) { | ||
| i++; | ||
| } else { | ||
| return [arr[i], arr[j]]; | ||
| } | ||
| } | ||
|
|
||
| return []; | ||
| } |
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