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" start point.
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" start point.
思路一:暴力求解,两层for循环搞定,但是效率较低,可AC,程序如下所示:
/**
* 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[] res = new int[len];
for (int i = 0; i < len; ++ i){
int bias = Integer.MAX_VALUE, index = 0;
for (int j = 0; j < len; ++ j){
if (i == j){
continue;
}
if (intervals[j].start >= intervals[i].end){
if (bias > intervals[j].start - intervals[i].end){
bias = intervals[j].start - intervals[i].end;
index = j;
}
}
}
res[i] = (bias == Integer.MAX_VALUE)?-1:index;
}
return res;
}
}
思路二:因为题目说明了start值无重复,因此:
1、通过一个hashMap记录start到数组下标的映射
2、先将intervals按照start值排序,如[3,4],[2,3],[1,2],排序后:[1, 2], [2, 3],[3, 4]
3、定义一层for循环,依次寻找各个区间对应的新的区间,如[1,2]对应区间为[2, 3],[2, 3]对应区间[3, 4], [3, 4]对应null
4、根据hashMap值,在找到原始数组对应的下标值
后续有时间补上算法。