leetcode436. Find Right Interval

题目要求

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.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

假设一个二维的整数数组中每一行表示一个区间，每一行的第一个值表示区间的左边界，第二个值表示区间的右边界。如今要求返回一个整数数组，用来记录每个边界右侧最邻近的区间。

如[ [1,4], [2,3], [3,4] ]表示三个区间，[1,4]不存在最邻近右区间，所以[1,4]的最邻近右区间的位置为-1。[2,3]最邻近右区间为[3,4]，所以返回2，[3,4]也不存在最临近右区间，所以也返回-1。最终函数返回数组[-1,2,-1]。

思路1：二分法

若是咱们将区间按照左边界进行排序，则针对每个区间的右边界，只要找到左边界比这个值大的最小左边界所在的区间便可。这里不可以直接对原来的二维数组进行排序，由于会丢失每个区间的原始下标位置。代码中采用了内部类Node来记录每个区间的左边界以及每个区间的原始下标，并对Node进行排序和二分法查找。代码以下：

public int[] findRightInterval(int[][] intervals) {        
        int[] result = new int[intervals.length];
        Arrays.fill(result, -1);
        
        Node[] leftIndex = new Node[intervals.length];
        for(int i = 0 ; i<intervals.length ; i++) {
            Node n = new Node();
            n.index = i;
            n.value = intervals[i][0];
            leftIndex[i] = n;
        }
        Arrays.sort(leftIndex);
        
        for(int i = 0 ; i<intervals.length ; i++) {
            int rightIndex = intervals[i][1];
            int left = 0, right = intervals.length-1;
            while(left <=right) {
                int mid = (left + right) / 2;
                Node tmp = leftIndex[mid];
                if(tmp.value > rightIndex) {
                    right = mid - 1;
                }else {
                    left = mid + 1;
                }
            }
            if(leftIndex[right].value == rightIndex) {
                result[i] = leftIndex[right].index;
            }else if(right<intervals.length-1) {
                result[i] = leftIndex[right+1].index;
            }
            
        }
        return result;
    }
    
    public static class Node implements Comparable<Node>{
        int value;
        int index;
        @Override
        public int compareTo(Node o) {
            // TODO Auto-generated method stub
            return this.value - o.value;
        }
    }

思路二：桶排序

换一种思路，有没有办法将全部的区间用一维数组的方式展开，一维数组的每个下标上的值，都记录了该位置所在的右侧第一个区间。具体流程以下：

1. 假设输入为[3,4], [2,3], [1,2]的三个区间，展开来后咱们知道这里的三个区间横跨了[1,4]这么一个区间
2. 咱们能够用长度为4的一维数组bucket来表示这个完整的区间，其中每个下标表明的位置都以左边界做为偏移值，如数组的0下标其实表明位置1，数组的1下标表明位置2。
3. 先根据全部区间的左值更新区间数组，如[3,4]表明着区间数组中位置为3-1=2的位置的第一个右侧区间为[3,4], 所以bucket[2]=0, 同理bucket[1]=0, bucket[0]=1
4. 开始查询时，若是当前桶下标上没有记录右侧的第一个区间，则不停的向右遍历，直到找到第一个值。若是没找到，则说明其右侧不存在区间了。反过来，也能够从右往左遍历bucket数组，每遇到一个没有填充的位置，就将右侧的值赋予当前位置。

代码以下：

public int[] findRightInterval(int[][] intervals) {
        int[] result = new int[intervals.length];
        int min = intervals[0][0], max = intervals[0][1];
        
        for(int i = 1 ; i<intervals.length ; i++) {
            min = Math.min(min, intervals[i][0]);
            max = Math.max(max, intervals[i][1]);
        }
        
        int[] buckets = new int[max - min + 1];
        Arrays.fill(buckets, -1);
        for(int i = 0 ; i<intervals.length ; i++) {
            buckets[intervals[i][0] - min] = i;
        }
        
        for(int i = buckets.length-2 ; i>=0 ; i--) {
            if(buckets[i] == -1) buckets[i] = buckets[i+1]; 
        }
        
        for(int i = 0 ; i<intervals.length ; i++) {
            result[i] = buckets[intervals[i][1] - min];
        }
        return result;
    }

这里的核心思路在于，如何理解bucket数组，这个bucket数组本质上是将全部的区间以最左边界和最右边界展开，数组的每个下标对应着区间中的相对位置，并记录了这个下标右侧的第一个区间的位置。