06 Leader

题目 1：Dominator

A zero-indexed array A consisting of N integers is given. Thedominator of array A is the value that occurs in more than half of the elements of A.

For example, consider array A such that
A[0] = 3 A[1] = 4 A[2] = 3A[3] = 2 A[4] = 3 A[5] = -1A[6] = 3 A[7] = 3

The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.

Write a function

int solution(vector<int> &A);

that, given a zero-indexed array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.

Assume that:

N is an integer within the range [0..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, given array A such that
A[0] = 3 A[1] = 4 A[2] = 3A[3] = 2 A[4] = 3 A[5] = -1A[6] = 3 A[7] = 3

the function may return 0, 2, 4, 6 or 7, as explained above.

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

int solution(vector<int> &A) {
// write your code in C++11
int n=A.size();
if(n==0) return -1;
int leader=A[0],countleader=1;

for(int i=1;i<n;i++)
{
if(A[i]!=leader)
{
countleader--;
if(countleader<1)
{
leader=A[i];
countleader=1;
}
}
else
countleader++;
}
countleader=0;
int index=0;
//
for(int i=0;i<n;i++)
{
if(A[i]==leader)
{
countleader++;
index=i;
}
}
if((countleader<<1)>n)
return index;
else
return -1;
}

题目 2：EquiLeader

A non-empty zero-indexed array A consisting of N integers is given.

The leader of this array is the value that occurs in more than half of the elements of A.

An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.

For example, given array A such that:
A[0] = 4 A[1] = 3 A[2] = 4 A[3] = 4 A[4] = 4 A[5] = 2

we can find two equi leaders:

0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.
2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4.

The goal is to count the number of equi leaders.

Write a function:

int solution(vector<int> &A);

that, given a non-empty zero-indexed array A consisting of N integers, returns the number of equi leaders.

For example, given:
A[0] = 4 A[1] = 3 A[2] = 4 A[3] = 4 A[4] = 4 A[5] = 2

the function should return 2, as explained above.

Assume that:

N is an integer within the range [1..100,000];
each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

int solution(vector<int> &A) {
// write your code in C++11
int n=A.size();
if(n==1) return 0;
int leader=A[0],countleader=1;
for(int i=1;i<n;i++)
{
if(A[i]!=leader)
{
countleader--;
if(countleader<1)
{
leader=A[i];
countleader=1;
}
}
else
countleader++;
}
countleader=0;
//find leader
for(int i=0;i<n;i++)
{
if(A[i]==leader)
{
countleader++;
}
}
if((countleader<<1)<=n)
return 0;
int equileader=0;
int leadernum=0;
for(int i=0;i<n;i++)
{
if(A[i]==leader)
{
leadernum++;
}
if(((leadernum<<1)>i+1)&&((countleader-leadernum)<<1)>n-i-1)
equileader++;

}

9803
return equileader;
}