s*****w
发帖数: 1527 | 1 say
Array 1 = {1, 3, 2, 4}, idx is 0, 1 ,2 3
Array 2 = {1, 2, 3, 4}
how do i know it's idx is actually 0, 2, 1, 3 from Array 1 ?
is there such function ? |
|
c**t
发帖数: 2744 | 2 var qry = from b in Array2
select Array.IndexOf(Array1, b);
say
Array 1 = {1, 3, 2, 4}, idx is 0, 1 ,2 3
Array 2 = {1, 2, 3, 4}
how do i know it's idx is actually 0, 2, 1, 3 from Array 1 ?
is there such function ? |
|
d****g
发帖数: 1049 | 3 这两天考古看见很多人推荐这个。去找了一下出来一堆。什么international idx,
spartan 500, spartan 500 idx advtg。大家说的是哪个啊？给个tick name吧。thx |
|
h*******i
发帖数: 717 | 4 在网上下了字幕，但后缀名是idx, 这个怎么打开呀？哪位高手给指点一下，多谢！ |
|
|
h*******i
发帖数: 717 | 6 对对对，是一个是idx, 一个是rar, 可还是不知道怎么打开阿？多谢多谢！ |
|
L*******g
发帖数: 913 | 7 解压rar得到sub文件，确定idx和sub的文件名和你的电影文件一样，直接播放电影就行
了。否则要手动在你的播放器里选择字幕文件。 |
|
u*********r
发帖数: 1181 | 8 上面部分是费用，下面部分是各自相应的average annual total return。
非常感谢啊
Gross Expense Ratio：
=======================
TIER I - LIFECYCLE FUNDS
- show details. - hide details.
LIFEPATH 2015 07/05/2006 Blended Fund Investments* N/A 0.1221
% No additional fees apply.
LIFEPATH 2020 06/30/2006 Blended Fund Investments* N/A 0.1092
% No additional fees apply.
LIFEPATH 2025 07/05/2006 Blended Fund Investments* N/A 0.1237
% No additional fees apply.
LIFEPATH 2030 06/3... 阅读全帖 |
|
n*********e
发帖数: 3 | 9 理解题意有误...如果不跳出的话，那就是16x16的矩阵
转移概率每次不变（不知对不对）.好像还是没有直接回答“空格子的数量的
stationary distribution”的问题。。。wondering
%transition matrix
A = zeros(16,16);
% corner/side
corner = [1 4 13 16];
side = [2 3 9 12;5 8 14 15];
for idx = 1:numel(corner)
A(corner(idx),side(:,idx)) = 0.5;
end
% side
side = side(:);
for idx = 1:numel(side);
row = ceil(side(idx) / 4);
col = rem(side(idx), 4);
left = side(idx) - 1;
right = side(idx) + 1;
up = side(idx) - 4;
down = side(idx) + 4;
if rem(left... 阅读全帖 |
|
z****u
发帖数: 104 | 10 C version，求指正
int NextNumber(int i)
{
int max_length = 1;
int* digits = (int*) malloc(sizeof(int) * max_length);
/* break number down to digits */
int idx = 0;
while(i)
{
if (idx > max_length - 1)
{
max_length *= 2;
digits = (int*) realloc(digits, sizeof(int) * max_length);
}
digits[idx++] = i % 10;
i /= 10;
}
int length = idx;
/* find the break digit */
for (idx = 1; idx < length; idx++)
... 阅读全帖 |
|
f*******t
发帖数: 7549 | 11 找出了一年多前写的逆波兰处理算数表达式的代码，强烈建议有兴趣的自己实现一下：
#include
#include
#include
#include
#define BUFFSIZE 1024
using namespace std;
struct Token {
bool isNum;
int num;
char op;
Token();
Token(const Token& t);
};
Token::Token()
{
isNum = false;
num = 0;
op = 0;
}
Token::Token(const Token& t)
{
isNum = t.isNum;
num = t.num;
op = t.op;
}
Token *getToken(const char expr[], int& idx)
{
Token *res = NULL;
while(expr[idx] == ' ')
... 阅读全帖
|
|
s****0
发帖数: 117 | 12 大概是这个意思. 我没理解最后两种情况怎么处理，好像结果不确定。
package myutil;
import java.util.Arrays;
public class FlexSearch {
final Integer[] data;
final boolean asc;
public FlexSearch(Integer[] data) {
if (data == null || data.length < 2)
throw new java.lang.NoSuchFieldError();// whatever.
this.data = data;
asc = data[0] - data[1] < 0;
}
private int getIdx(int val) {
return Arrays.binarySearch(data, val);
}
public int getLTMax(int val) {
int... 阅读全帖 |
|
J**9
发帖数: 835 | 13 How about this direction?
=====================================
/// array is sorted in Ascending order
int binarySearchAscending(int array[], int key, int low, int high)
{
int mid;
if(!array|| (low>high))
return -1;
while(high >= low)
{
mid = low+(high-low)/2;
if(key < array[mid])
high = mid-1;
else if (key > array[mid])
low = mid+1;
else
return mid;
}
return -1;
}
/// array is sorted in D... 阅读全帖 |
|
C*****n
发帖数: 1049 | 14 我觉得可能是思路的问题吧，楼主把问题看得复杂了，有点儿机械性地从脑海里搜索各
种数据结构，觉得有些像就先用上，实际可能完全没必要。
比如Bob和David是同事那题，哪里用得上并查集，他提示的那个类里用个list就能解决
问题。
又比如side walk那个题，哪里需要merge interval，就定义一个
struct interv {
double left=0, right=0.01;
bool isWet() {
return left>=right;
}
};
就好了，然后生成0,1之间的一个均匀分布，模拟每个雨滴看落到哪个interval里，直
到100个都打湿。
vector sidewalk(100,interv());
int cnt=0, wetCnt=0, idx;
while(wetCnt<100) {
++cnt;
double p= (double)(rand())/RAND_MAX;
double left=p-0.005;
... 阅读全帖 |
|
i**********e
发帖数: 1145 | 15 写了一个 递归+cache，这题还可以再优化。
int cache[512];
int solve(const string &num, int idx) {
if (cache[idx] != -1) return cache[idx];
if (idx == num.length()) return 1;
if (num[idx] == '0') return 0;
if (idx == num.length()-1) return 1;
int count = solve(num, idx+1);
if (num[idx] == '1' || (num[idx] == '2' && num[idx+1] <= '6'))
count += solve(num, idx+2);
cache[idx] = count;
return count;
}
int solve(const string &num) {
for (int ... 阅读全帖 |
|
h***k
发帖数: 161 | 16 把二楼的c++版翻译了一下，能过但应该还能优化
public class Solution {
/**
*@param A: A positive integer which has N digits, A is a string.
*@param k: Remove k digits.
*@return: A string
*/
public String DeleteDigits(String A, int k) {
// write your code here
int len = A.length();
if (len == k) {
return "";
}
int idx = 0;
int[] num = new int[len];
for (int i = 0; i < len; i++) {
num[i] = A.charAt(i) - '0';
... 阅读全帖 |
|
h***k
发帖数: 161 | 17 把二楼的c++版翻译了一下，能过但应该还能优化
public class Solution {
/**
*@param A: A positive integer which has N digits, A is a string.
*@param k: Remove k digits.
*@return: A string
*/
public String DeleteDigits(String A, int k) {
// write your code here
int len = A.length();
if (len == k) {
return "";
}
int idx = 0;
int[] num = new int[len];
for (int i = 0; i < len; i++) {
num[i] = A.charAt(i) - '0';
... 阅读全帖 |
|
S******t
发帖数: 151 | 18 我贴一个第一题能通过的代码吧：
vector bestA;
int bestLen;
void search(int idx, int sum, int len, vector>>& f,
vector& ret, vector& A) {
//cout << idx << " " << sum << " " << len << endl;
if (idx == 0) {
int lenA = bestA.size();
vector v = ret;
/*
for (int i = 0; i < v.size(); i++)
cout << v[i] << " ";
cout << endl;
*/
if (v.size() < lenA || lenA == 0) {
bestA = v;
return;
... 阅读全帖 |
|
n*********e
发帖数: 3 | 19 Just a thought. Please correct if this is not right.
If use Markov transition matrix, we may calculate the probability that a
mouse jump from cell i to j after N iterations.
The transition matrix A is built as follows:
There are 36 status. 4x4 of them are the board; 20 of then are the outside
cells around the board. The transition matrix is a 36x36 matrix.
One element in the matrix A, A_ij means the probability at a time that a
mouse jumps from cell i to j, where i,j range from 1 to 36.
You can ... 阅读全帖 |
|
r*********n
发帖数: 4553 | 20 solution to Q1
std::vector find3idx( const std::vector& a ){
std::vector idx;
int sz = a.size();
if( sz == 0 ) return idx;
idx.push_back(0);
for(int i = 1; i < sz; i++){
if( a[i] < a[idx.back()] ) {
idx.back() = i;
continue;
}else if( a[i] > a[idx.back()] ){
idx.push_back(i);
if( idx.size() == 3 ) return idx;
}
}
return idx;
}
I assume it is the user's responsibility to check whethe... 阅读全帖 |
|
a********e
发帖数: 53 | 21 这种用dfs的时间复杂度，总是绕不清楚。
code如下，
分析时间复杂度哪种比较对？
一种是dfs一共执行了n！次，每次有n^2, 所以总共是O(n!n^2).
另一种是T(n)= nT(n-1) + n^2, 所以是? 我也不知道了。
////////////////////////////////////
bool PosOK(int col[], int idx){
for ( int i=0; i< idx; i++){
if (col[i]== col[idx] || (idx-i)== abs(col[idx] - col[i] ) )
return false;
}
return true;
}
void dfs2(int col[], int idx, vvs & res, int n){
//when there is a solution
if (idx==n){
string s(n, '.');
vs tmp(n, s);
for (int i=0; i< n; i++)
tmp[i][co... 阅读全帖 |
|
b********g
发帖数: 28 | 22 Here is the code:
public class URLAverage{
private static final int MAX_LEN = 400;
private int total = 0;
private int sumCount = 0;
private double avg = 0.0;
private int idx = 0;
private int offset = 0;
int[] count = new int[MAX_LEN];
public double getAverage(){
if(offset + sumCount == 0) return 0.0;
return (sumCount * avg + offset * (idx + 1)) / (offset + sumCount);
}
public void addURL(String url){
int k = url.length();
count[k - 1]++;
if(k - 1 < idx){
... 阅读全帖 |
|
s******7
发帖数: 1758 | 23 我试着写了下，用java bitset会简单点，assume Location里有block的index, 这种题
没见过估计和楼主一样死翘翘
class Location{
int idx; //block index
public Location(int idx){
this.idx=idx;
}
}
public class myPBuffer extends pBuffer{
BitSet bitSet = new BitSet(1024);
@Override
public Location allocate() throws RuntimeException {
return new Location(bitSet.nextClearBit(0));
}
@Override
public void put(Location l, byte[] data) {
int p = l.idx*1024;
int len = Math.m... 阅读全帖 |
|
k****r
发帖数: 807 | 24 some logic as follows:
helper(s, 0, 0, "", res, new HashSet set);
public void helper(String s, int idx, int blc, String tmp, List res,
Set set) {
if (idx == s.length()) {
if (blc == 0 && !set.contains(tmp)) {
res.add(tmp);
set.add(tmp);
}
return;
}
if (s.charAt(idx) == '(') {
helper(s, idx + 1, blc + 1, tmp + "(", res);
helper(s, idx + 1, blc + 1, tmp, res);
} else if (s.charAt(idx) == ')') ... 阅读全帖 |
|
r********g
发帖数: 1351 | 25 写了个递归的，复杂度。。应该是C(n,k)吧？指数级的？
class Solution {
public:
vector > combinationSum2(vector &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
sort(num);
vector > Results;
vector tmpR;
doFind(num, Results, 0, target, tmpR);
return Results;
}
void doFind(vector &num, vector > &Results, int idx,
int target, vector tmpR) {
i... 阅读全帖 |
|
l*********8
发帖数: 4642 | 26 贴个代码吧：
int scanPostOrder(int *A, int idx, int minVal, int maxVal)
{
if (idx < 0 || A[idx] < minVal || A[idx] > maxVal)
return idx; //如果不能满足数值范围,或者已经扫描完毕，就返回到父节点
int val = A[idx];
idx = scanPostOrder(A, idx-1, val, maxVal); // 右子树
return scanPostOrder(A, idx, minVal, val); // 左子树
}
bool checkPostOrder(int *A, int n)
{
return scanPostOrder(A, n-1, INT_MIN, INT_MAX) < 0;
} |
|
l*******0
发帖数: 63 | 27 这不是针对fresh grad的吧？不懂设计模式啊。。。
不过第二题你的基本思路是对的。要想少一点空间的话，可以把string先转成char
array. FYI:
char[] p=str.toCharArray();
static void dfs(char[] p, int idx, ArrayList sol){
if(idx==p.length){ //find one string
sol.add(new String(p));
}else{ //replace ? to either 0 or 1
if(p[idx]=='?'){
p[idx]='0';
dfs(p,idx+1,sol);
p[idx]='1';
dfs(p,idx+1,sol);
... 阅读全帖 |
|
l****x
发帖数: 87 | 28 static void TestFunc()
{
string s = "WXYZwxyz";
Encode(s,261);
cout<
}
static void Encode(string& s, int n)
{
int offSet = n%26;
int idx = -1;
int tmp = 0;
while (++idx < s.size())
{
tmp = s[idx] + offSet;
if(s[idx] >= 'A' && s[idx] <= 'Z')
s[idx] = tmp > 'Z'?(tmp - 1) % 'Z' + 'A':tmp;
else if(s[idx] >= 'a' && s[idx] <= 'z')
s[idx] = tmp >... 阅读全帖 |
|
p****6
发帖数: 3 | 29 this is the c++ version, based on greedy method.
string DeleteDigits(string A, int k) {
// wirte your code here
int len = A.size();
if(len==k) return "0";
int idx = 0;
while(k>0){
if(A[idx]>A[idx+1]){
A.erase(A.begin()+idx);
k--;
} else{
while(A[idx] <= A[idx+1])
idx++;
A.erase(A.begin()+idx);
k--;
idx = 0;
... 阅读全帖 |
|
p****6
发帖数: 3 | 30 this is the c++ version, based on greedy method.
string DeleteDigits(string A, int k) {
// wirte your code here
int len = A.size();
if(len==k) return "0";
int idx = 0;
while(k>0){
if(A[idx]>A[idx+1]){
A.erase(A.begin()+idx);
k--;
} else{
while(A[idx] <= A[idx+1])
idx++;
A.erase(A.begin()+idx);
k--;
idx = 0;
... 阅读全帖 |
|
M******a
发帖数: 6723 | 31 http://www.ideobook.com/doc/chinese_academia_corruption.pdf
亦 明
一 前言
学术腐败是九十年代之后风靡中国大陆的一种社会现象，它是继政治腐败和经济腐败之
后的另一个大规模、深层次的社会腐败。学术腐败的具体表现就是，学术界人士，上至
院士、
博导，下至研究生、大学生，抄袭剽窃成风，巧取豪夺成性，弄虚作假为常，欺世盗名
为荣。
不仅如此，学术腐败已经从学者的个体行为发展成集体、集团行为，并且有制度化、合
理化
的趋势。比如，大学普遍向高官明赠暗送高等学位，学术界头面人物公开地、明目张胆
地欺
骗政府、舆论和社会，学术机构对学术腐败现象不仅熟视无睹，任其泛滥，甚至包庇纵
容，
等等。
不过，在中国，对学术腐败现象的讨论和揭露目前主要停留在互联网上。在社会上，中
国的一般民众对学术腐败现象几乎一无所知，或不甚关心。在大学，在科研机构，尽管
人们
承认学术腐败的存在，但对它的严重程度却看法截然不同。在中国的常规媒体上，讨论
学术
腐败问题的文章和报道充其量也就是零星的，表面的。而中国的政府部门至今未对学术
腐败
问题表明自己的态度，或... 阅读全帖 |
|
D********g
发帖数: 650 | 32 加上了backtracking:
static class DPElement {
int value;
int prevCostIdx; // 0-based cost idx
int takenThisElement; // 0/1, indicating whether taking current
element, idx 0 based.
public DPElement() {
value = 0;
prevCostIdx = 0;
takenThisElement = 0;
}
}
static String word2Anagram(final String word) {
if (word == null) {
return null;
}
int ht[] = new int[256];
f... 阅读全帖 |
|
D********g
发帖数: 650 | 33 我的code：
先扫一遍找到A的个数和B的个数
然后再扫一遍处理A
再扫一遍处理B
code里处理A和B的算法不太一样，第二种更好一点。另外，有没有包子？
/*input is assumed to end at '\0' and input is assumed to be long enough to
hold the
* transformed string
* */
static char[] deleteADoubleB(final char[] input) {
if (input == null || input.length == 0) {
return input;
}
int aCount = 0, bCount = 0, originalLen = 0;
for (int i = 0; i < input.length; ++i) {
if (input[i] == 'A') {
aCount ... 阅读全帖 |
|
e****e
发帖数: 418 | 34 凑个热闹!
public int calculate( String[] in, int sIdx ) {
if ( sIdx == in.length - 1)
return toInt( in[sIdx] );
int addOrMinus = getAddOrMinusSign( in[sIdx + 1] );
if ( addOrMinus != 0 ) {
if ( sIdx + 2 == in.length - 1 || getAddOrMinusSign( in[sIdx + 3] ) !=
0 ) {
in[sIdx+2] = toStr( toInt( in[sIdx] ) + addOrMinus*toInt( in[sIdx+
2] ) );
return calculate( in, sIdx + 2 );
} else {
calMulOrDivAndSetBack( in, sIdx+2 );
in[sIdx+3] = in[s... 阅读全帖 |
|
e****e
发帖数: 418 | 35 凑个热闹!
public int calculate( String[] in, int sIdx ) {
if ( sIdx == in.length - 1)
return toInt( in[sIdx] );
int addOrMinus = getAddOrMinusSign( in[sIdx + 1] );
if ( addOrMinus != 0 ) {
if ( sIdx + 2 == in.length - 1 || getAddOrMinusSign( in[sIdx + 3] ) !=
0 ) {
in[sIdx+2] = toStr( toInt( in[sIdx] ) + addOrMinus*toInt( in[sIdx+
2] ) );
return calculate( in, sIdx + 2 );
} else {
calMulOrDivAndSetBack( in, sIdx+2 );
in[sIdx+3] = in[s... 阅读全帖 |
|
w***y
发帖数: 6251 | 36 要怎么做才能不超时？
我就是按照传统的N queens写法， 但是大集合到12就超时了
public class Solution {
int res=0;
public int totalNQueens(int n) {
int[] row = new int[n];
res=0;
_solveNQueens(row,0);
return res;
}
public void _solveNQueens(int[] row, int idx){
//when it finishes the last row, output the valid result
if(idx==row.length){
res++;
return;
}
... 阅读全帖 |
|
f**********t
发帖数: 1001 | 37 Well, my code is complex and sigfault. Let me go back if I have time.
This is too hard for a 45 minutes interview.
void skipSpace(const char **s) {
while (isspace(**s)) {
++(*s);
}
}
int getNum(const char **s) {
bool neg = false;
if (**s == '-') {
neg = true;
++(*s);
}
int cur = 0;
while (isdigit(**s)) {
cur = cur * 10 + (**s - '0');
++(*s);
}
--(*s);
return neg ? -cur : cur;
}
int opNum(int x, int y, char op) {
switch (op) {
case '+':
return x + y... 阅读全帖 |
|
t**r
发帖数: 3428 | 38 我的代码：
很糟糕，大数据过不去
应该可以用bitmap 优化一下 今天懒得弄了
package topcoder;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;
//B[0] = 0
//B[i] = A[B[i-1]], for every i between 1 and N-1, inclusive.
/*
* Permutation Child array
{0, 1, 2} {0, 0, 0}
{0, 2, 1} {0, 0, 0}
{1, 0, 2} {0, 1, 0}
{1, 2, 0} {0, 1, 2}
{2, 0, 1} {0, 2, 1}
{2, 1, 0} {0, 2, 0}
* */
public... 阅读全帖 |
|
r****7
发帖数: 2282 | 39 不熟java或者c＃，但是用c＋＋我觉得可以写的很简洁，要是不想reverse string的话
用反过来的iterator来做while循环的判断，不过不太记得怎么写的夜懒得放狗搜了.
reverse string的cost反正也不大
string stringAdd(string l, string r) {
string result;
int carrier = 0;
size_t idx = 0;
reverse(l.begin(), l.end());
reverse(r.begin(), r.end());
while (carrier != 0 || idx < l.size() || idx < r.size()) {
int currValue = carrier + idx < l.size() ? l[idx] - '0' : 0 + idx <
r.size() ? r[idx] - '0' : 0;
if (currValue >= 10) {
c... 阅读全帖 |
|
S******6
发帖数: 55 | 40 一道很简单的leetcode 题目： Reverse Words in a String III
我的解法（c#）一直被报 Memory Limit Exceeded 错误。谁知道如何改进？ 多谢了先
。：）
public class Solution {
public string ReverseStr(string str)
{
if(str.Length < 2) return str;
int i = str.Length-1;
string resS = null;
while(i >= 0)
{
resS += str[i--];
}
return resS;
}
public string ReverseWords(string s) {
string result = null;
if(s.Length < 2) return s;
int i = 0;
... 阅读全帖 |
|
t*********r
发帖数: 63 | 41 别理2楼的，哪有人天生什么都会。多刷题，多问问题，多看文章，多写代码，自然就
会提高了。
我对c#不熟悉，但是看你的算法没啥问题，估计就是string拷贝的问题了。我改了一下
，用stringbuilder代替string，并且减少了不必要的拷贝，你看看吧。
public class Solution {
public string ReverseWords(string s) {
StringBuilder result = new StringBuilder();
if(s.Length < 2) return s;
int i = 0;
int idxS = 0;
while(i < s.Length)
{
while( idxS < s.Length && s[idxS] != ' ')
{
idxS++;
}
... 阅读全帖 |
|
|
