[Leetcode] 715. Range Module 解题报告

题目：

A Range Module is a module that tracks ranges of numbers. Your task is to design and implement the following interfaces in an efficient manner.

• addRange(int left, int right) Adds the half-open interval [left, right), tracking every real number in that interval. Adding an interval that partially overlaps with currently tracked numbers should add any numbers in the interval [left, right) that are not already tracked.

• queryRange(int left, int right) Returns true if and only if every real number in the interval [left, right) is currently being tracked.

• removeRange(int left, int right) Stops tracking every real number currently being tracked in the interval [left, right).

  Example 1:
  

  addRange(10, 20): null
  removeRange(14, 16): null
  queryRange(10, 14): true (Every number in [10, 14) is being tracked)
  queryRange(13, 15): false (Numbers like 14, 14.03, 14.17 in [13, 15) are not being tracked)
  queryRange(16, 17): true (The number 16 in [16, 17) is still being tracked, despite the remove operation)
  

  Note:

• A half open interval [left, right) denotes all real numbers left <= x < right.
• 0 < left < right < 10^9 in all calls to addRange, queryRange, removeRange.
• The total number of calls to addRange in a single test case is at most 1000.
• The total number of calls to queryRange in a single test case is at most 5000.
• The total number of calls to removeRange in a single test case is at most 1000.

  思路：

  本题目可以用线段树来实现，但是这里我们用二分查找树来实现。我们定义一个map<int, int> invals，用来表示互不相交的线段，例如invales[left] = right就可以表示一个左闭右开的区间[left, right)。下面就看看各个函数分别怎么实现：
  1）addRange(int left, int right)：首先找出[left, right)的前一个interval，然后判断我们是否需要在插入interval之前合并已有的interval，如果需要，则首先将待合并的interval删除，并更新left和right，最后将[left, right)插入二分查找树中。该函数的时间复杂度是O(nlogn)。

  2）queryRange(int left, int right)：首先找出小于等于[left, right)的一个interval，然后判断它的右边是否覆盖掉了[left, right)，如果是，则返回false，否则说明该interval覆盖掉了[left, right)，所以返回true。该函数的时间复杂度仍然是O(nlogn)。

  3）removeRange(int left, int right)：和addRange类似，我们首先找出和[left, right)重合的interval(s)，然后将该区间内的interval(s)都移除掉。但是需要注意的是，如果左边有残余的interval或者右边有残余的interval，则需要分别将它们保留。该函数的时间复杂度仍然是O(nlogn)。

  整个算法的空间复杂度是O(n)。

  代码：

  class RangeModule {
  public:
      RangeModule() {
          
      }
      
      void addRange(int left, int right) {
          auto l = invals.upper_bound(left), r = invals.upper_bound(right); 
          if (l != invals.begin()) {      // find the iterator that is before the new segment
              l--;
              if (l->second < left) {
                  l++;
              }
          }
          if (l != r) {                   // if l != r, we should merge the intervals
              left = min(left, l->first);
              right = max(right, (--r)->second);
              invals.erase(l,++r);
          }
          invals[left] = right;
      }
      
      bool queryRange(int left, int right) {
          auto it = invals.upper_bound(left);
          if (it == invals.begin() || (--it)->second < right) {
              return false;
          }
          return true;
      }
      
      void removeRange(int left, int right) {
          auto l = invals.upper_bound(left), r = invals.upper_bound(right); 
          if (l != invals.begin()) {
              l--;
              if (l->second < left) {
                  l++;
              }
          }
          if (l == r) {                   // no interval to remove
              return;
          }
          int l1 = min(left, l->first), r1 = max(right, (--r)->second);
          invals.erase(l, ++r);
          if (l1 < left) {                // a smaller interval left in the left
              invals[l1] = left;
          }
          if (r1 > right) {               // a smaller interval left in the right
              invals[right] = r1;
          }
      }
  private:
      map<int, int> invals;
  };
  
  
  /**
   * Your RangeModule object will be instantiated and called as such:
   * RangeModule obj = new RangeModule();
   * obj.addRange(left,right);
   * bool param_2 = obj.queryRange(left,right);
   * obj.removeRange(left,right);
   */