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436. Find Right Interval

扶摇_hyber 2022-08-03 阅读 54


Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the “right” of i.

For any interval i, you need to store the minimum interval j’s index, which means that the interval j has the minimum start point to build the “right” relationship for interval i. If the interval j doesn’t exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

Note:
You may assume the interval’s end point is always bigger than its start point.
You may assume none of these intervals have the same start point.
Example 1:

Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right"

Example 3:

Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right"

思路:
把给定区间的右边界与这些区间中的左边界进行比较,选择出,左边界比给定区间右边界大,且距给定区间最近区间。获取该区间的位置。
利用java TreeMap的性质,把所有区间的左边界作为key值,所在位置作为value值,保存在map中,利用TreeMap中已有的ceilingKey(key k)方法,直接获取与给定key大且最近的key值,然后通过key值得到位置。

/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
class Solution {
public int[] findRightInterval(Interval[] intervals) {
int len = intervals.length;
int nums[] = new int[len];
TreeMap<Integer,Integer> map = new TreeMap<Integer,Integer>();
for(int i = 0; i < len; i++){
map.put(intervals[i].start, i);
}

for(int i = 0; i < len; i++){
Integer num = map.ceilingKey(intervals[i].end);
if(num == null){
nums[i] = -1;
}else{
nums[i] = map.get(num);
}
}
return nums;
}
}


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