x*******9
发帖数: 138 | 1 做了一下
>> 给定一个数字数组 ,其中每个元素是从末端数小于原数组中该元素的个数。求原数
组。
Time complexity: O(n * logn) // quick-sort + traverse
Memo complexity: O(n)
>> 令s[i]为a[i+1..n-1]中比a[i]大的数的数量。求最大的s[i]。要求O(nlogn)
Time complexity: O(n * logn) // Manipulate the Binary Indexed Tree
Memo compleixty: O(n)
#include
#include
#include
#include
#include
#include
using namespace std;
#define print(x) cout << x << endl
#define input(x) cin >> x
const int SIZE = 1024;
// @brief Bin... 阅读全帖 |
|
b****e
发帖数: 45 | 2 多谢!确实是个很低级的错误...
不过我觉得这个办法修改一下还是可行的。按照你的提醒，更新的思路如下：
1. 对每一个值，都判断是否大于first并且小于second, 如果是则更新second为当前值;
2. 用一个新的min_val变量记录当前为止遇到的最小值，遇到increased pair时，如果
不能返回结果，则更新first使其为min_val, 并继续扫描。
更新后的代码如下，请指正~
vector FindIncreasingNum(vector vec) {
if (vec.size() < 3)
return vector();
int first = INT_MAX;
int second = INT_MAX;
int min_val = vec[0];
for (int i = 1; i < vec.size(); i++) {
if (vec[i] > vec[i - 1]) {
if (vec[i] > second) {
... 阅读全帖 |
|
w****x
发帖数: 2483 | 3 /*
Given tweet's inverted index，how to find phrases combination，e.g
phrase "twitter good tool", "twitter is a good tool" is better than "twitter
is good,
facebook is a better tool"
*/
bool GetClosestPhrase(hash_map>& dic, vector& strs,
int& nStart, int& nEnd)
{
for (vector::iterator it = strs.begin(); it != strs.end(); it++)
{
if (dic.find(*it) == dic.end())
return false;
}
int nNum = strs.size();
vector*> vec;
ve... 阅读全帖 |
|
w****x
发帖数: 2483 | 4 /*
Given tweet's inverted index，how to find phrases combination，e.g
phrase "twitter good tool", "twitter is a good tool" is better than "twitter
is good,
facebook is a better tool"
*/
bool GetClosestPhrase(hash_map>& dic, vector& strs,
int& nStart, int& nEnd)
{
for (vector::iterator it = strs.begin(); it != strs.end(); it++)
{
if (dic.find(*it) == dic.end())
return false;
}
int nNum = strs.size();
vector*> vec;
ve... 阅读全帖 |
|
s****e
发帖数: 638 | 5 来自主题: JobHunting版 - G新鲜面经 1.2 暴力加剪枝
vector used;
void transform(vector A, vector& vec){
int i=vec.size()-1;
if (i>0) {
if (( i&1 && vec[i] < vec[i-1]) ||
(!(i&1) && vec[i] > vec[i-1])) return;
}
if (vec.size()==A.size()) {
for(int i = 0 ; i < vec.size() ; ++i) cout<
cout<
return;
}
for(int i = 0 ; i < A.size() ; ++i) {
if (!used[i]) {
used[i] = true;
vec.push_back(A[i]);
transform(A, vec);
vec.pop_back();
used[i] = fa... 阅读全帖 |
|
h****y
发帖数: 137 | 6 double find_kth_largest_unknown_length(istringstream &sin, int k)
{
vector vec;
double x;
while (sin >> x)
{
vec.emplace_back(x);
if (vec.size() == (k << 1) - 1)
{
nth_element(vec.begin(), vec.begin() + k - 1, vec.end(), greater
());
vec.resize(k);
}
}
nth_element(vec.begin(), vec.begin()+k-1, vec.end(), greater());
return vec[k-1];
} |
|
h****y
发帖数: 137 | 7 double find_kth_largest_unknown_length(istringstream &sin, int k)
{
vector vec;
double x;
while (sin >> x)
{
vec.emplace_back(x);
if (vec.size() == (k << 1) - 1)
{
nth_element(vec.begin(), vec.begin() + k - 1, vec.end(), greater
());
vec.resize(k);
}
}
nth_element(vec.begin(), vec.begin()+k-1, vec.end(), greater());
return vec[k-1];
} |
|
f*******d
发帖数: 339 | 8 一般说来，如果总电荷不为零，通过选择坐标有可能使有限系统的电偶极矩为零， 因为
\vec p = \int \vec x \rho(x)
如果不为零， 可以取
\vec x' = \vec x - \vec x0
\vec p' = \int \vec x \rho (x) - \vec x0 \int rho (x)
所以只要选
\vec x0 = (\int rho(x) \vec x)/\int \rho(x) = （\int rho(x) \vec x） / Q_tot
就足以令电偶极矩为零。这里的条件是总电荷不为零， 如果总电荷为零就没有办法了。
电四极矩有五个独立分量，独立的坐标变换有六个（三个平移三个旋转）， 因此也可以
选座标
使之为零， 具体公式我就不写了，总之是写出其变换方程，然后求解。
但是不能同时使电偶极距为零。 如果总电荷和电偶极矩都恰为零， 那么上述平移变换不
起作用，
也无法使电四极矩为零。 总之， 最低一级不为零的多极矩是与坐标无关的，不可能消去
。 |
|
s******e
发帖数: 108 | 9 可以用 hashset 来替代 vector ,
public class RandomHash {
HashMap > hashmap = new HashMap
LinkedHashSet>();
Vector vec = new Vector();
public void insert(K key) {
vec.add(key);
if( ! hashmap.containsKey(key) ) {
LinkedHashSet hashSet = new
LinkedHashSet();
hashSet.add(new Integer(vec.size()-1));
hashmap.put(key,hashSet);
} else {
h... 阅读全帖 |
|
g*****x
发帖数: 799 | 10 void rangeBinarySearch(const vector &vec, const int val, int &range_beg
, int &range_end)
{
range_beg = -1;
range_end = -1;
int beg = 0;
int end = vec.size();
int mid;
while(beg < end)
{
mid = (end - beg) / 2 + beg;
if(vec[mid] == val && (mid == 0 || vec[mid - 1] < val))
{
range_beg = mid;
break;
}
if(vec[mid] >= val)
end = mid;
else
beg = mid + 1;
}
if(range_b... 阅读全帖 |
|
S******0
发帖数: 71 | 11 改成BFS了
//each time can jump up or down 2^k steps, find fewest jumping step,
//always start from the first step
vector getJumpSteps(int nCur, int nMaxStep)
{
vector vec;
if (nCur <= 0 || nMaxStep < nCur)
return vec;
int flg = 1;
while (nCur + flg <= nMaxStep)
{
vec.push_back(nCur + flg);
flg = (flg << 1);
}
flg = 1;
while (nCur - flg > 0)
{
vec.push_back(nCur - flg);
flg = (flg << 1);
}
return vec;
}
int... 阅读全帖 |
|
r*c
发帖数: 167 | 12 //矩阵求连接图
#include
#include
#include
#include
#include
using namespace std;
enum ColorEnum{White, Black};
struct Cell
{
int row;
int col;
Cell(int r, int c): row(r), col(c){}
};
struct Cluster{
bool visited;
vector vec;
Cluster(int i, int j) : visited(false) {
vec.push_back(Cell( i, j));
}
bool isWithinRange(const Cell& a, const Cell& b){
return abs(a.row - b.row) <= 1 ... 阅读全帖 | |
|
r*c
发帖数: 167 | 13 //矩阵求连接图
#include
#include
#include
#include
#include
using namespace std;
enum ColorEnum{White, Black};
struct Cell
{
int row;
int col;
Cell(int r, int c): row(r), col(c){}
};
struct Cluster{
bool visited;
vector vec;
Cluster(int i, int j) : visited(false) {
vec.push_back(Cell( i, j));
}
bool isWithinRange(const Cell& a, const Cell& b){
return abs(a.row - b.row) <= 1 ... 阅读全帖 | |
|
b*******t
发帖数: 459 | 14 比如我要定义一个class叫做vec（向量），它有三个变量和一堆函数：
class vec {
public:
double x,y,z;
public:
vec(){};
~vec() {};
...
}
现在我希望定义某种union结构能让我也能用vec.xa[0],vec.xa[1],vec.xa[2]
来访问vec.x, vec.y, vec.z，如何能够实现呢？谢谢。。。 |
|
s******e
发帖数: 108 | 15 public class RandomHash {
HashMap hashmap = new HashMap();
Vector vec = new Vector();
public void insert(K key) {
vec.add(key);
hashmap.put(key, new Integer(vec.size()-1));
}
public K getRand() {
if(vec.size()<=0)
return null;
Random randomGenerator = new Random();
int rndnum = randomGenerator.nextInt(vec.size());
return vec.get(rndnum);
}
public boole... 阅读全帖 |
|
l*****a
发帖数: 14598 | 16 void GetPath(int size, int x,int y,vector& vec)
{
if(x==size&&y==size)
{ vector::iterator it;
for(it=vec.begin();it!=vec.end();++it) { cout<<*it; }
cout<
return;
}
if(x
vec.push_back('R');
GetPath(size,x+1,y,vec);
vec.pop_back();
}
if(y
vec.push_back('D');
GetPath(size,x,y+1,vec);
vec.pop_back();
}
return;
} |
|
w****x
发帖数: 2483 | 17 迭代法:
//Find the nth number which is composed by factor 2, 3 and 5
void PrintSerials(int n)
{
assert(n > 0);
vector vec;
vec.push_back(1);
int nI2 = 0;
int nI3 = 0;
int nI5 = 0;
int nCount = 1;
while(nCount < n)
{
int nTmp2 = vec[nI2]*2;
int nTmp3 = vec[nI3]*3;
int nTmp5 = vec[nI5]*5;
int nMin = min(min(nTmp2, nTmp3), nTmp5);
if (vec.back() != nMin)
{
vec.push_back(nMin);
nCount++;
... 阅读全帖 |
|
p*****3
发帖数: 488 | 18 const int N = 100;
void printWays(int dp[N][N], int i, int j, vector& vec)
{
if (i < 0 || j < 0 || dp[i][j] <= 0)
return;
if (j == 0)
{
for (vector::iterator it = vec.begin(); it != vec.end(); it++)
cout<<*it<<" ";
cout<
return;
}
vec.push_back(i);
printWays(dp, i, j-i, vec);
vec.pop_back();
printWays(dp, i-1, j, vec);
}
int getUniqueWays(int n)
{
if (n <= 0)
return 0;
int dp[N][N] = { 0 }... 阅读全帖 |
|
m**q
发帖数: 189 | 19 同样，当检查一个新的点时，需要把x-h之前的点从map中去掉,
其实可以用一个map存[x-h,x]中的点，key是y坐标。在向map中
添加一个新的点时，先检查map里是否有y坐标相同的点，有则删除；
同时把x-h之前的点依据它们的y坐标，把它们从map中删除。
总复杂度不超过O(nlgn)。
#include
#include |
|
x*******5
发帖数: 152 | 20 谢谢大牛指导，但是好像C++的还是不work
void Pearl::GetSubset(vector& array, int target,int start,vector&
vec){
if (vec.size()==target){
Pearl::Print(vec);
return;
}
if (start==array.size()) return;
for (int i=start;i
vec.push_back(array[i]);
GetSubset(array,target,start+1,vec);
vec.pop_back();
}
return;
}
测试：
int a[]={1,2,3};
vector v(a,a+3);
vector vec;
Pearl::GetSubset(v,2,0,vec);
Outp... 阅读全帖 |
|
w****x
发帖数: 2483 | 21 做了无数recursion的题后花了两小时才整出gray code解, 大家鄙视我吧, 哈哈哈:
/*
The gray code is a binary numeral system where two successive values differ
in only one bit.
Given a non-negative integer n representing the total number of bits in the
code, print the sequence of gray code. A gray code sequence must begin with
0.
For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:
00 - 0
01 - 1
11 - 3
10 - 2
*/
void _inner_sol(int& nCur, vector& vec, int nStopPos, int nPos)
{
if (nPos < 0)
{
... 阅读全帖 |
|
b****e
发帖数: 45 | 22 刚写了个第一题的代码：
vector FindIncreasingNum(vector vec) {
int first = INT_MAX;
int second = INT_MAX;
for (int i = 1; i < vec.size(); i++) {
if (vec[i] > vec[i - 1]) {
if (vec[i] > second) {
vector res;
res.push_back(first);
res.push_back(second);
res.push_back(vec[i]);
return res;
}
else {
first = vec[i - 1];
second = vec[... 阅读全帖 |
|
w****x
发帖数: 2483 | 23 那个烙印这些题根本就是想灭你嘛， 楼主不要自责。
中国人最后那题可能是要这个解法：
void inner_get_rightmost(NODE* pNode, vector& vec, int nLev)
{
if (NULL == pNode)
return;
if (nLev >= vec.size())
vec.push_back(pNode);
inner_get_rightmost(pNode->pRgt, vec, nLev+1);
inner_get_rightmost(pNode->pLft, vec, nLev+1);
}
vector getRightMost(NODE* pRoot)
{
vector vec;
inner_get_rightmost(pRoot, vec, 0);
return vec;
} |
|
w****x
发帖数: 2483 | 24 vector getJumpSteps(int nCur, int nMaxStep)
{
vector vec;
if (nCur <= 0 || nMaxStep < nCur)
return vec;
int flg = 1;
while (nCur + flg <= nMaxStep)
{
vec.push_back(nCur + flg);
flg = (flg << 1);
}
flg = 1;
while (nCur - flg > 0)
{
vec.push_back(nCur - flg);
flg = (flg << 1);
}
return vec;
}
int getMinStep(int n, int nMaxStep)
{
if (n <= 0 || nMaxStep < n)
return -1;
int* rec = new int[nM... 阅读全帖 |
|
d**********x
发帖数: 4083 | 25 应该没问题，和coldknight的程序在2-10000之内做了对比，结果完全一致。他的
nMaxStep我设置为了2 * target，不知道会不会引起问题。
在纸面上也做了不太严格的证明。。这样复杂度就是O(logn)。
对比程序：（前半部分还是coldknight的）
#include
#include
#include
using namespace std;
vector getJumpSteps(int nCur, int nMaxStep)
{
vector vec;
if (nCur <= 0 || nMaxStep < nCur)
return vec;
int flg = 1;
while (nCur + flg <= nMaxStep)
{
vec.push_back(nCur + flg);
flg = (flg << 1);
}
flg = 1;
while (nCur - ... 阅读全帖 |
|
l**n
发帖数: 67 | 26 I am confused here and don't know you are asking or answering.
Doesnot distinguish partial deriv. and deriv. is very bad notation.
But just to remind you, you are dealing with a field equation
here.(mass conservation) Whenever you want to use the field equation,
all the variables should be expressed in terms of field variables
which are coordinates \vec{x} and time. So
\rho=\rho(\vec{x},t)
\vec{u}=\vec{u}(\vec{x},t)
\vec{a}=\vec{a}(\vec{x},t)
and you have the \partial\rho/\partial t + div(\rho u |
|
f*********5
发帖数: 576 | 27 void GetPath(node * root, vector&vec,int target)
{
if(!root) return;
if(vec.size()>0)
{ sum=0;
for(int i=vec.size()-1;i>=0;i--)
{ sum+=vec[i];
if(sum==target) { PRINT;break;}
}
}
vec.push_back(root->data);
if (root->left) GetPath(root->left,vec,target);
if (root->right) GetPath(root->right,vec,target);
return;
}
level
Design an |
|
w****x
发帖数: 2483 | 28 struct NODE
{
int nVal;
NODE* pLft;
NODE* pRgt;
NODE(int x) : nVal(x), pLft(NULL), pRgt(NULL) {}
};
struct TreeIter
{
stack m_stk;
void movNext()
{
if (m_stk.empty())
return;
NODE* pCur = m_stk.top();
if (pCur->pRgt != NULL)
{
pushLft(pCur->pRgt);
return;
}
while (!m_stk.empty() && pCur != m_stk.top()->pLft)
{
pCur = m_stk.top();
m_stk.pop... 阅读全帖 |
|
M*****M
发帖数: 20 | 29 来自主题: JobHunting版 - 一道面试题 类似找最大BST subtree， bottom up.
bool findsubtree(TreeNode *node, unordered_set &set, vector
&vec) {
if(not node) {
return true;
}
unordered_set leftset;
unordered_set rightset;
bool left = findsubtree(node->left, leftset, vec);
bool right = findsubtree(node->right, rightset, vec);
if(left&&right) {
//leftset and rightset overlap
for(auto ele:leftset) {
if(rightset.find(ele)!=rightset.end()) {
... 阅读全帖 |
|
x*******9
发帖数: 138 | 30
should
5
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define print(x) cout << x << endl
#define input(x) cin >> x
string make_result(const vector& vec, int idx) {
string res = "";
int len = vec.size();
for (int i = 0; i < idx; i++) {
res += to_string(vec[i]);
}
res += "(";
for (int i = idx; i < len; i++) {
... 阅读全帖 |
|
x******a
发帖数: 6336 | 31 在visual studio express 2012里为什么compile不过去？
#include
#include
#include
#include
int main()
{
std::vector vec{3, 4, 2, 9, 15, 267};
for(auto& elem:vec) elem*=3;
std::ostream_iterator out_it(std::cout, ", ");
std::copy(vec.begin(), vec.end(), out_it);
return 0;
}
Error 1 error C2601: 'vec' : local function definitions are illegal
c:users。。documentscodetest11test11main.cpp 8 1 test11
Error 2 error C2143: synt... 阅读全帖 |
|
s***e
发帖数: 911 | 32
In my derivation, all the quantities
are field quantities which must be functions of the space coordinates and the
time. What we need to do now is to express the speed field q(\vec{x},t) by
\vec{x}(\vec{a},t).
In your picture, \vec{x}(\vec{a},t) should be the so-called streamline. A
given \vec{a} correspods to a steamline, and all the possible
\vec{a} give the field. We know that there is no intersecting streamlines
in the field, so a space coordinate only belongs to a streamline.
Then we have |
|
f*********5
发帖数: 576 | 33 //assume that a[]={1,5,10,25} already sorted
void GetCombination(int a[],int size,int target,vector& vec)
{
if(target==0) {cout<< <
if(target<=0) return;
for(int i=0,i
{
if(vec.size()>0)
{
if(a[i]
}
if (a[i]>target) break;
vec.push_back(a[i]);
GetCombination(a,size,target-a[i],vec);
vec.pop_back();
}
return;
} |
|
x***y
发帖数: 633 | 34 another way to use backtracking
//find all the commbination
typedef vector::size_type VecIndex;
void find_all_comb ( unsigned target, vector & vec, vector
>& eleCount, VecIndex vIndex){
//vec is assumed sorted
if(vIndex==vec.size())
return ;
unsigned mul = target/vec[vIndex];
eleCount[vIndex] = mul;
if(!(target%vec[vIndex])){
for(VecIndex vi=0; vi
for(VecIndex vc=0; vc
cout<< vec[vi] << |
|
w****x
发帖数: 2483 | 35 struct NODE
{
vector vecGUID;
NODE* nodes[256];
NODE() { memset(nodes, 0, sizeof(nodes)); }
};
void _inner_get_guid(NODE* pNode, const char* str, int k, vector& vec)
{
if (NULL == pNode)
return;
if (k <= 0)
{
vec = pNode->vecGUID;
return;
}
_inner_get_guid(pNode->nodes[*str], str+1, k-1, vec);
}
vector getGUID(const char* str, NODE* pRoot, int k)
{
vector vecRet;
if (NULL == pRoot || NULL == str || *str == 0 ... 阅读全帖 |
|
w****x
发帖数: 2483 | 36 //serialize, deserialize vector
int serialize(vector& vec, char* str)
{
if (str == NULL)
return 0;
char* pWriter = str;
*((int*)pWriter) = vec.size();
pWriter += sizeof(int);
for (int i = 0; i < vec.size(); i++)
{
strcpy(pWriter, vec[i].c_str());
pWriter += 1 + strlen(pWriter);
}
return pWriter - str;
}
void deserialize(char* str, vector& vec)
{
if (NULL == str)
return;
int nSize = *((int*)str);
c... 阅读全帖 |
|
d**********x
发帖数: 4083 | 37 实际上是O(logn)的空间，但是既然长度为2^32的数列可以借助一个int来作为辅助空间
，这点overhead基本可以忽略不计了。欢迎找bug：
#include
#include
#include
using namespace std;
#include
static int get_index(int i) {
if (i == 0) {
return 1;
}
return (1 << i) + 1;
}
void rearrange(vector& vec) {
int bitmap = 0; //"O(1)" space for length < (1 << 32)
int k = vec.size() / 2;
int n = vec.size();
for (int i = 0; get_index(i) < k; ++i) {
int cur = get_index(i);
... 阅读全帖 |
|
p*****2
发帖数: 21240 | 38 (defn f [vec]
(defn- dfs [vec start p len]
(let [next (nth vec p)]
(if (= next start)
(inc len)
(dfs vec start next (inc len)))))
(apply max (for [i (range 0 (count vec))]
(dfs vec i i 0)))) |
|
p*****2
发帖数: 21240 | 39 (defn f [vec]
(defn- dfs [vec start p len]
(let [next (nth vec p)]
(if (= next start)
(inc len)
(dfs vec start next (inc len)))))
(apply max (for [i (range 0 (count vec))]
(dfs vec i i 0)))) |
|
G****A
发帖数: 4160 | 40 来自主题: JobHunting版 - 一道面试题 void increment(vector& vec)
{
int pos = vec.size() - 1;
uint carry = 1;
do{
if (pos < 0)
vec.insert(vec.begin(), 1);
else{
int sum = vec[pos] + carry;
vec[pos] = sum % 10;
carry = sum / 10;
pos--;
}
} while(carry);
}
that |
|
c**********x
发帖数: 32 | 41 int getArithSeq(vector & vec){
if (vec.size() < 3)
return 0;
int start = 0, i=2, ret=0;
for (; i < vec.size(); ++i){
if (vec[i - 1] * 2 == vec[i] + vec[i - 2]){
if (start == 0)
start = i - 2;
}
else{
ret += cal(start, i-1);
start = 0;
}
}
if (start != 0)
ret += cal(start, i - 1);
return ret;
} |
|
x*******9
发帖数: 138 | 42 SPOJ MDOLLS
4A（有点蠢
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define print(x) cout << x << endl
#define input(x) cin >> x
int n;
vector > vec;
int main() {
int T;
int a, b;
input(T);
while (T--) {
input(n);
vec.clear();
multiset st;
for (int i = 0; i < n; i++) {
input(a >> b);
vec.push_back({a, b})... 阅读全帖 |
|
D**u
发帖数: 204 | 43 Here is a solution by using vector calculations.
Let a = |BC|, c = |AB|, then easy to calculate that:
(1) |AM|^2 = |MB|^2 + |AB|^2 = a^2 -2ac + 2c^2
(2) |CN|^2 = |CB|^2 + \NB|^2 = 2a^2 - 4ac + 4c^2.
We also have
vec(MA)\cdot vec(CN) = (vec(BM)-vec(BA))\cdot (vec(BC)-vec(BN))
= a^2 - 2ac + 2c^2 = cos(pi/4) * |AM| * |CN|.
so 角MPC=45度 |
|
X***9
发帖数: 7385 | 44
哦？你很了解 动量守恒定律？动量定理？和其区别？关系？
根据wiki解释
动量具有一个特殊属性：只要是在一个封闭系统中，它总会保持恒定，即使是物体碰撞
发生时。而对动能而言，非弹性碰撞的物体的动能将不会守恒。因此，当碰撞过后可利
用动量守恒来计算未知速度。
在物理学上，这个特殊属性被用来来解决两个相碰物体的问题。因为动量始终保持恒定
，碰撞前动量的总和一定与碰撞后动量的总和相等：
m_1 vec{v}_{1,i} + m_2 vec{v}_{2,i} = m_1 vec{v}_{1,f} + m_2 vec{v}_
{2,f}，
其中，i表示碰撞前的初量，f表示碰撞后的末量。要注意的是此時vec{v}為向量。
通常来说，我们只需知道碰撞前（或碰撞后）物体的速度便可计算出碰撞后（或碰撞前
）物体的速度。要正确解决这个问题意味着你必须知道将会发生怎样的碰撞。碰撞有两
种类型，两种类型中动量都保持恒定：
大哥
飞鸟撞飞机是非弹性碰撞
你不能否定非弹性碰撞和动量守恒的关系。 |
|
D***h
发帖数: 183 | 45 how about this. I test it seems ok.
#include
#include
using namespace std;
void num2alpha(int n){
vector vec;
vector::reverse_iterator it;
int i=n;
if(n==0)
vec.push_back('a');
else{
while(i>0){
vec.push_back('a'+n%26);
i=n/26;
n = n/26 -1;
}
}
for(it=vec.rbegin();it!=vec.rend();it++)
cout<<*it;
cout<
} |
|
l*****a
发帖数: 14598 | 46 void GetCombination(int a[],int size,int start,vector &vec)
{
for(int i=start;i<=size-1;i++)
{
if((i!=start)&&(a[i]==a[i-1])) continue;
vec.push_back(a[i]);
PrintCombination(vec);
if(i
vec.pop_back();
}
} |
|
l*****a
发帖数: 14598 | 47 void GetCombination(int a[],int size,int start,vector &vec)
{
for(int i=start;i<=size-1;i++)
{
if((i!=start)&&(a[i]==a[i-1])) continue;
vec.push_back(a[i]);
PrintCombination(vec);
if(i
vec.pop_back();
}
} |
|
l*****a
发帖数: 14598 | 48 void GetSubset(T array[],size_t size,int start,size_t target,vector&vec)
{
if (vec.size()==target) {PRINT; return;}
if(start ==size) return;
for(int i=start;i
{
vec.push_back(array[i]);
GetSubset(array,size,i+1,target,vec);
vec.pop_back();
}
return;
} |
|
w****x
发帖数: 2483 | 49 /*
Serialize/DeSerialize a tree
*/
struct NODE
{
int nVal;
vector vec;
NODE(int n) : nVal(n) {}
};
void _inner_serial(NODE* pNode, char*& p)
{
if (NULL == pNode)
return;
*p++ = pNode->vec.size();
*p++ = pNode->nVal;
for (vector::iterator it = pNode->vec.begin();
it != pNode->vec.end(); it++)
_inner_serial(*it, p);
}
const char* Serialize(NODE* pRoot, char mem[])
{
if (NULL == mem || NULL == pRoot)
return NULL;
char... 阅读全帖 |
|
w****x
发帖数: 2483 | 50 //Get all trees according to preorder
struct NODE
{
int nVal;
NODE* pLft;
NODE* pRgt;
NODE(int n) : nVal(n), pLft(NULL), pRgt(NULL) {}
};
void GetComb(int a[], int n, vector& vec)
{
if (n <= 0)
return ;
if (n == 1)
{
vec.push_back(new NODE(a[0]));
return;
}
vector vecTmp1;
GetComb(a+1, n-1, vecTmp1);
for (vector::iterator it = vecTmp1.begin();
it != vecTmp1.end(); it++)
{
NODE* pRoot = ne... 阅读全帖 |
|
