Find Right Interval

Find Right Interval

Description

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.

Tags: Hash Table

解讀題意

給出一組區間（無序），現在要在這組區間裏，找到每個區間的右區間（可能有多個），能夠滿足右區間的start要大於等於當前區間的end。找到右區間中start值最小的，並返回其在數組中的下標。

思路1

用一個數據結構來存儲區間start與下標之間的映射關係，TreeMap本身有排序的性質（logn），現有它來存儲映射關係。ps：用到了TreeMap的一個api，ceilingEntry：smallest entry whose key > given key
1. 將intervals[i].start作爲key，下標i爲value存入TreeMap中，爲區間的start和數組下標建立映射關係。
2. 遍歷區間數組intervals[]，用當前的intervals[i].end,通過ceilingEntry方法，找到在TreeMap中，比intervals[i].end大且在TreeMap中最小的。若存在，則返回當前entry的value，即下標；若不存在，則返回-1

public class Solution {

     public int[] findRightInterval(Interval[] intervals) {
        int[] result = new int[intervals.length];
        TreeMap<Integer, Integer> intervalMap = new TreeMap<>();

        for (int i = 0; i < intervals.length; ++i) {
            intervalMap.put(intervals[i].start, i);
        }

        for (int i = 0; i < intervals.length; ++i) {
            Map.Entry<Integer, Integer> entry = intervalMap.ceilingEntry(intervals[i].end);
            result[i] = (entry != null) ? entry.getValue() : -1;
        }

        return result;
    }
}

time complexity：O(nlogn)

leetCode彙總：https://blog.csdn.net/qingtian_1993/article/details/80588941

項目源碼，歡迎star：https://github.com/mcrwayfun/java-leet-code