题目链接:https://leetcode.com/problems/find-right-interval/description/
题目大意:给定一组区间(i, j),找出每一个区间“右边”最近的区间的索引
思路:
将题目给的interval 组转化为带有索引的indexIntreval组, 然后排序,接着对每个区间,遍历其右边的区间比较
代码:
/**
* 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:
struct indexInterval {
int start;
int end;
int index;
indexInterval():indexInterval(-1,-1,-1){}
indexInterval(int s, int e, int i): start(s), end(e), index(i) {}
};
bool static compare(indexInterval a, indexInterval b) {
return a.start <= b.start;
}
vector<int> findRightInterval(vector<Interval>& intervals) {
vector<indexInterval> localSet;
int len = intervals.size();
for(int i = 0; i < len; i++) {
int start = intervals[i].start;
int end = intervals[i].end;
localSet.push_back(indexInterval(start, end, i));
}
sort(localSet.begin(), localSet.end(), compare);
vector<int> result(len,-1);
for(int i = 0; i < len; i++) {
int target = -1;
for(int j = i + 1; j < len; j++) {
if(localSet[j].start >= localSet[i].end) {
target = localSet[j].index;
break;
}
}
result[localSet[i].index] = target;
// target = -1;
}
return result;
}
};