LeetCode 436. Find Right Interval 题解(C++)
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LeetCode 436. Find Right Interval 题解(C++)
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.
补充
- 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.
举例
Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.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;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.
思路
- 首先用map保存每个Interval的起点及其对应的位置,key保存的是Interval的start,value保存的是该Interval的位置i;
- 之后遍历intervals,对每一个Interval,在hashMap中查找key值大于等于Interval.end,且两者差值与Interval.end最接近的Interval(用map的lower_bound函数实现),并返回其迭代器ite,若ite等于hashMap.end(),则表示找不到大于等于Interval.end的结点,此时将-1保存到数组里,若ite不等于hashMap.end(),则表示找到了符合要求的结点,将其value值(即其位置)保存到数组里。
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: vector<int> findRightInterval(vector<Interval>& intervals) { int len = intervals.size(); vector<int> result; map<int, int> hashMap; for (int i = 0; i < len; ++i) { hashMap[intervals[i].start] = i; } for (auto interval : intervals) { auto ite = hashMap.lower_bound(interval.end); if (ite == hashMap.end()) { result.push_back(-1); } else { result.push_back(ite->second); } } return result; }};
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