436. Find Right Interval （寻找最近的右区间）

谁践踏了优雅 2022-07-14 08:37 31阅读 0赞

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:

1.  You may assume the interval's end point is always bigger than its start point.
2.  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.

题目大意：给定一些区间，找到每个区间的最近右区间，要保证右区间的start要大于等于当前区间的end，返回值为最近右区间在数组中的下标，如果某个区间不存在右区间，即该区间的end比所有区间的最大的start还要大，则返回-1。

解题思路：先复制一份给定的区间数组，命名为sortedIntervals\[\]，在该副本中，修改区间的end属性值，用以记录该区间在原数组中的下标，然后对该副本数组sortedIntervals\[\]按照start值从小到大进行排序。然后遍历原数组intervals\[\]，对于其中的每一个区间对象，通过二分查找在排好序的sortedIntervals\[\]数组中查找最近的start值（要求在右边），然后返回对应的区间对象在原始数组中的下标值（现在存放在end属性中），对于end值比sortedIntervals\[\]中最大的start值还大的对象，返回-1。

代码如下（本以为效率很低，没想到提交之后发现22ms，beats 93.65%）

/**
     * 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) {
    		int n = intervals.length;
    		int[] res = new int[n];
    		int index = 0;
    		Interval[] sortedIntervals = new Interval[n];
    		//复制原始数组，用end值存放该区间对象在原始数组中的下标值
    		for(Interval interval : intervals){
    			sortedIntervals[index] = new Interval(interval.start, index);
    			index++;
    		}
    		
    		List<Interval> list = new ArrayList<>(Arrays.asList(sortedIntervals));
    		Collections.sort(list, new Comparator<Interval>() {
    
    			@Override
    			public int compare(Interval o1, Interval o2) {
    				return o1.start - o2.start;
    			}
    		});
    		
    		sortedIntervals = (Interval[]) list.toArray(new Interval[n]);
    		int max = sortedIntervals[n - 1].start;
    		index = 0;
    		
    		for (Interval interval : intervals) {
    			if (interval.end > max)
    				res[index] = -1;
    			else
    				res[index] = binarySearch(sortedIntervals, interval.end);
    			index++;
    		}
    		return res;
    	}
    
    	public int binarySearch(Interval[] intervals, int target) {
    		int left = 0, right = intervals.length - 1;
    		int mid;
    		while (left <= right) {
    			mid = (left + right) / 2;
    			if (intervals[mid].start == target) {
    				return intervals[mid].end;
    			} else if (intervals[mid].start < target) {
    				left = mid + 1;
    			} else {
    				right = mid - 1;
    			}
    		}
    		return intervals[left].end;
    	}
    }