题目描述:

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:

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Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

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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:

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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.

使用排序+二分搜索. 要注意记录排序之前每个interval的下标.

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/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<int> findRightInterval(vector<Interval>& intervals) {
vector<pair<int, int>> maps(intervals.size());
for(int i = 0; i < intervals.size(); i++){
maps[i].first = intervals[i].start;
maps[i].second = i;
}

sort(maps.begin(), maps.end(), [&](pair<int, int> &a, pair<int, int> &b){
return a.first < b.first;
});

vector<int> ans(intervals.size());
for(int i = 0; i < intervals.size(); i++){
int target = intervals[i].end;
ans[i] = binSearch(maps, 0, target);
}
return ans;
}

int binSearch(vector<pair<int, int>> &m, int begin, int target){
int left = begin, right = m.size(), mid = (left + right) / 2;
while(left < right){
if(m[mid].first == target){
return m[mid].second;
}
else if(m[mid].first < target){
left = mid + 1;
}
else{
right = mid;
}
mid = (left + right) / 2;
}
return -1;
}
};

或者可以使用hash表.

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/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<int> findRightInterval(vector<Interval>& intervals) {
unordered_map<int, int> maps;
for(int i = 0; i < intervals.size(); i++){
maps[intervals[i].start] = i;
}
vector<int> ans(intervals.size());
for(int i = 0; i < intervals.size(); i++){
ans[i] = maps.count(intervals[i].end) ? maps[intervals[i].end] : -1;
}
return ans;
}
};