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.
这道题给了我们一堆区间，让我们找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，由于区间的顺序不能变，所以我们不能给区间排序，我们需要建立区间的 start 和该区间位置之间的映射，由于题目中限定了每个区间的 start 都不同，所以不用担心一对多的情况出现。然后我们把所有的区间的 start 都放到一个数组中，并对这个数组进行降序排序，那么 start 值大的就在数组前面。然后我们遍历区间集合，对于每个区间，我们在数组中找第一个小于当前区间的 end 值的位置，如果数组中第一个数就小于当前区间的 end，那么说明该区间不存在右区间，结果 res 中加入-1；如果找到了第一个小于当前区间 end 的位置，那么往前推一个就是第一个大于等于当前区间 end 的 start，我们在 HashMap 中找到该区间的坐标加入结果 res 中即可，参见代码如下：
解法一：
class Solution {
public:
vector<int> findRightInterval(vector<vector<int>>& intervals) {
vector<int> res, starts;
unordered_map<int, int> m;
for (int i = ; i < intervals.size(); ++i) {
m[intervals[i][]] = i;
starts.push_back(intervals[i][]);
}
sort(starts.rbegin(), starts.rend());
for (auto interval : intervals) {
int i = ;
for (; i < starts.size(); ++i) {
if (starts[i] < interval[]) break;
}
res.push_back((i > ) ? m[starts[i - ]] : -);
}
return res;
}
};
上面的解法可以进一步化简，我们可以利用 STL 的 lower_bound 函数来找第一个不小于目标值的位置，这样也可以达到我们的目标，参见代码如下：
解法二：
class Solution {
public:
vector<int> findRightInterval(vector<vector<int>>& intervals) {
vector<int> res;
map<int, int> m;
for (int i = ; i < intervals.size(); ++i) {
m[intervals[i][]] = i;
}
for (auto interval : intervals) {
auto it = m.lower_bound(interval[]);
if (it == m.end()) res.push_back(-);
else res.push_back(it->second);
}
return res;
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/436
类似题目：
Data Stream as Disjoint Intervals
参考资料：
https://leetcode.com/problems/find-right-interval/
https://leetcode.com/problems/find-right-interval/discuss/91819/C%2B%2B-map-solution