LeetCode 436. Find Right Interval 题解（C++）

LeetCode 436. Find Right Interval 题解（C++）

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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;
    }
};