436 Find Right Interval 寻找右区间
Description:
You are given an array of intervals, where intervals[i] = [start_i, end_i] and each start_i is unique.
The right interval for an interval i is an interval j such that start_j >= end_i and start_j is minimized.
Return an array of right interval indices for each interval i. If no right interval exists for interval i, then put -1 at index i.
Example:
Example 1:
Input: intervals = [[1,2]]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: intervals = [[3,4],[2,3],[1,2]]
Output: [-1,0,1]
Explanation: There is no right interval for [3,4].
The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3.
The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.
Example 3:
Input: intervals = [[1,4],[2,3],[3,4]]
Output: [-1,2,-1]
Explanation: There is no right interval for [1,4] and [3,4].
The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.
Constraints:
1 <= intervals.length <= 2 * 10^4
intervals[i].length == 2
-10^6 <= starti <= endi <= 10^6
The start point of each interval is unique.
题目描述:
给定一组区间，对于每一个区间 i，检查是否存在一个区间 j，它的起始点大于或等于区间 i 的终点，这可以称为 j 在 i 的“右侧”。
对于任何区间，你需要存储的满足条件的区间 j 的最小索引，这意味着区间 j 有最小的起始点可以使其成为“右侧”区间。如果区间 j 不存在，则将区间 i 存储为 -1。最后，你需要输出一个值为存储的区间值的数组。
注意：
你可以假设区间的终点总是大于它的起始点。
你可以假定这些区间都不具有相同的起始点。
示例 :
示例 1:
输入: [ [1,2] ]
输出: [-1]
解释:集合中只有一个区间，所以输出-1。
示例 2:
输入: [ [3,4], [2,3], [1,2] ]
输出: [-1, 0, 1]
解释:对于[3,4]，没有满足条件的“右侧”区间。
对于[2,3]，区间[3,4]具有最小的“右”起点;
对于[1,2]，区间[2,3]具有最小的“右”起点。
示例 3:
输入: [ [1,4], [2,3], [3,4] ]
输出: [-1, 2, -1]
解释:对于区间[1,4]和[3,4]，没有满足条件的“右侧”区间。
对于[2,3]，区间[3,4]有最小的“右”起点。
思路:
对区间的左端点进行排序, 排序之后可以用二分查找查询
时间复杂度O(nlgn), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
vector findRightInterval(vector>& intervals)
{
map m;
vector result;
for (int i = 0; i < intervals.size(); i++) m[intervals[i].front()] = i;
for (auto interval : intervals) result.emplace_back(m.lower_bound(interval.back()) == m.end() ? -1 : m.lower_bound(interval.back()) -> second);
return result;
}
};
Java:
class Solution {
public int[] findRightInterval(int[][] intervals) {
int count[][] = new int[intervals.length][2], n = intervals.length, result[] = new int[intervals.length], m = 0;
for (int i = 0; i < n; i++) {
count[i][0] = intervals[i][0];
count[i][1] = i;
}
Arrays.sort(count, new Comparator(){ public int compare (int[] a, int[] b) { return a[0] - b[0]; } });
for (int interval[] : intervals) {
int target = interval[1], l = 0, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if (count[mid][0] < target) l = mid + 1;
else r = mid;
}
result[m++] = (l == n ? -1 : count[l][1]);
}
return result;
}
}
Python:
class Solution:
def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
sorted_intervals, result = sorted([(interval[0], i) for i, interval in enumerate(intervals)]), []
for interval in intervals:
target, l, r = interval[1], 0, len(intervals)
while l < r:
mid = (l + r) >> 1
if sorted_intervals[mid][0] < target:
l = mid + 1
else:
r = mid
result.append(-1 if l == len(intervals) else sorted_intervals[l][1])
return result