LeetCode 715. Range Module

原题链接在这里：https://leetcode.com/problems/range-module/

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.

题解：

Have a TreeMap to store existing interval, key is start of interval and value is end.

When adding range, get floor key of left, if key exists and its corresponding value >= left, then there is a merge, left = preKey.

get floor key of right, if key exists and it sorrsonding value >= right, then there is another merge, right = tm.get(nextKey).

Add the new range and clear the keys from (left, right].

When querry Range, just check if there is existing interval coverring [left, right].

When removing range, first get floor key of right, if it exists and value > right, then after removing, there is still some range left. Add new range [right, tm.get(nextKey)).

Do the similar to left. But we need to do right first, because we may change value of preKey if we do left first.

Then remove keys in range [left, right).

Time Complexity: addRange, O(nlogn). clear takes O(nlogn). queryRange, O(logn). removeRange, O(nlogn).

Space: O(n).

AC Java:

class RangeModule {
    TreeMap<Integer, Integer> tm;
    
    public RangeModule() {
        tm = new TreeMap<>();    
    }
    
    public void addRange(int left, int right) {
        if(left >= right){
            return;
        }
        
        Integer preKey = tm.floorKey(left);
        if(preKey != null && tm.get(preKey) >= left){
            left = preKey;
        }
        
        Integer nextKey = tm.floorKey(right);
        if(nextKey != null && tm.get(nextKey) >= right){
            right = tm.get(nextKey);
        }
        
        tm.put(left, right);
        tm.subMap(left, false, right, true).clear();
    }
    
    public boolean queryRange(int left, int right) {
        Integer preKey = tm.floorKey(left);
        if(preKey == null){
            return false;
        }
        
        return tm.get(preKey) >= right;
    }
    
    public void removeRange(int left, int right) {
        if(left >= right){
            return;
        }
        
        Integer nextKey = tm.floorKey(right);
        if(nextKey != null && tm.get(nextKey) > right){
            tm.put(right, tm.get(nextKey));
        }
        
        Integer preKey = tm.floorKey(left);
        if(preKey != null && tm.get(preKey) > left){
            tm.put(preKey, left);
        }
        
        tm.subMap(left, true, right, false).clear();
    }
}

/**
 * Your RangeModule object will be instantiated and called as such:
 * RangeModule obj = new RangeModule();
 * obj.addRange(left,right);
 * boolean param_2 = obj.queryRange(left,right);
 * obj.removeRange(left,right);
 */

类似Merge Intervals.