Colidity-- EquiLeader

由于要满足equi-leader，则必须是整个数组的leader，证明如下：leader > i+1/2 && leader > (n-i-1)/2，那么leader > n/2

我之前c++写了，但是有错五

static int equiLeader(int[] a){
  int len = a.length;
  int equi_leaders = 0;

  // first, compute the leader
  int leader = a[0];
  int ctr = 1;

  for(int i = 1; i < a.length; i++){
    if(a[i] == leader) ctr++;
    else ctr--;
    if(ctr == 0){
      ctr = 1;
      leader = a[i];
    }
  }

  // check if it's a leader?
  int total = 0;
  for(int i : a){
    if(i == leader) total++; 
  }

  if(total <= (len/2)) return 0; // impossible

  int ldrCount = 0;
  for(int i = 0; i < a.length; i++){
    if(a[i] == leader) ldrCount++;
    int leadersInRightPart = (total - ldrCount);
    if(ldrCount > (i+1)/2   &&   leadersInRightPart > (len-i-1)/2){
      equi_leaders++;
    }
  }

  return equi_leaders;
}