Known binary search tree and two values, find the number of nodes in the binary search tree, where the value of node is between two values

int foo(tree* node, int min, int max, int& count)
{
if(!root) return;
if(root->val>min && root->val < max)
{
count++;
foo(node->left,min,max,count);
foo(node->right,min,max,count);
}
else if(root->val>max) foo(node->left,min,max,count);
else if(root->val < min) foo(node->right,min,max,count);
}

The return value of the assignment operator is always *this instead of void, because assignment operator supports concatenation assignment like a=b=c

The return value of the assignment operator is always *this instead of void, because assignment operator supports concatenation assignment like a=b=c
http://en.wikipedia.org/wiki/Assignment_operator_in_C%2B%2B

Balance partition Input: give you a integer set {A[0]…A[n-1]}, 0 < A[i] < k, partition to two subsets S1 and S2, minimize |sum(S1)-sum(S2)|

Balance partition
Input: give you a integer set {A[0]…A[n-1]}, 0 < A[i] < k, partition to two subsets S1 and S2, minimize |sum(S1)-sum(S2)|
p(i,j)=1 means there is at least one subset in set i{A[0]…A[i-1]} has sum of j. i is from 0~n-1, j is from 0~(n-1)k
p(i,j)=0 means there is no subset in set i{A[0]…A[i-1]} has sum of j
so
p(i,j) =max{p(i-1,j),p(i-1,j-A[i])}
and define s=sum{A[0]…A[n-1]}/2
so minimize |sum(S1)-sum(S2)|= minimize |s-i:p(i,j)=1|

Longest increasing subsequence in array A[n]

Longest increasing subsequence in array A[n]
L(j)=max{L(i)}+1, where i < j and A[i] < A[j]
max_g=max{L(j)}
O(n^2)

int foo(int A[], int len)
{
int max_g=0;
int L[len];
for(int j=0;j < len;j++)
{
for(i=0; i < j; i++)
{
if(A[i] < A[j])
L[j]=max(L[j],L[i]+1);
}
max_g=max(max_g,L[j]);
}
return max_g;
}

Convert string to integer num, convert integer num to string

Convert string to integer num, convert integer num to string
1 A
2 B
…
27 AA
28 AB
int foo(char* str)
{
if(!str) return 0;
int num=0;
int len=strlen(str)-1;
int idx=len;
while(idx>=0)
{
num+=(str[idx]-'A'+1)*pow(26,len-idx);
idx--;
}
return num;
}
char* foo(int num)
{
if(num<=0) return NULL;
char* str=new char[11];
int idx=0;
while(num>=0)
{
str[idx]=num%26+'A'-1;
num/=26;
idx++;
}
str[idx]='\0';
return reverse(str);
}