436. Find Right Interval

http://www.cnblogs.com/EdwardLiu/p/6139796.html

https://leetcode.com/problems/find-right-interval/#/solutions

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.

Solution 1: TreeMap,      Time complexity: O(NlogN)

像这种在一个集合里面寻找有没有比某个数小的数，一般要么treeMap要么treeSet。Interval的题经常需要用treeMap, Data Stream as Disjoint Intervals 就是

This implementation provides guaranteed log(n) time cost for the containsKey, get, put and remove operations. 

map.higherEntry(key) 找到的是一个entry with least key strictly greater than the given key, 所以20行要减一，因为interval.start可能跟current interval.end重合

用map.cellingEntry(key)找的就是一个entry with least key greater than or equal to the given key

map.lowerKey(key) Returns the greatest key strictly less than the given key, or null if there is no such key.

map.floorKey(key) Returns the greatest key less than or equal to the given key, or null if there is no such key.

关键是找键值对和根据什么遍历, 会根据数组构造类, 会用map.higherEntry(key)

/**
 * 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; }
 * }
 */
public class Solution {
    public int[] findRightInterval(Interval[] intervals) {
        if(intervals==null || intervals.length==0) return null;
        int[] res = new int[intervals.length];
        TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
        for (int i=0; i<intervals.length; i++) {
            map.put(intervals[i].start, i);
        }
        for (int i=0; i<intervals.length; i++) {
            Interval cur = intervals[i];
            Map.Entry<Integer, Integer> entry = map.higherEntry(cur.end-1);
            if (entry != null) {
                res[i] = entry.getValue();
            }
            else res[i] = -1;
        }
        return res;
    }
}

　　

 Map.Entry接口
　　Map的entrySet()方法返回一个实现Map.Entry接口的对象集合。集合中每个对象都是底层Map中一个特定的键/值对。

　　通过这个集合的迭代器，您可以获得每一个条目(唯一获取方式)的键或值并对值进行更改。当条目通过迭代器返回后，除非是迭代器自身的remove()方法或者迭代器返回的条目的setValue()方法，其余对源Map外部的修改都会导致此条目集变得无效，同时产生条目行为未定义。
　　(1) Object getKey(): 返回条目的关键字
　　(2) Object getValue(): 返回条目的值
　　(3) Object setValue(Object value): 将相关映像中的值改为value，并且返回旧值