算法攻关-双指针笔记

right = left + K;
//链表倒数第K个节点类问题
while(right.next != null){
  left= left.next;
  }
  return left;
//链表循环检测问题 right比left快一步
right = left + K ; K =1;
while(left != right){
   if(left ==null || right.next == null){
     return false;
   }
   left = left.next;
   right = right.next.next;
}

when（right - left + 1 > minWindow）{
    //固定左指针，移动右指针
    if(condition1){
        //处理使其满足condition2条件,并不符合condition1条件
        right++;
    }
    //固定右指针，移动左指针
    if(condition2){
        //处理使其满足condition1条件,并不符合condition2条件
        left++;
    }
}