LeetCode 56. Merge Intervals C++ 比较器 与 Python 两种语言

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

# input:
[[1,3],[2,6],[8,10],[15,18]]
[[1,4],[4,5]]
[[1,4],[2,3]]
[[2,3],[4,5],[6,7],[8,9],[1,10]]
[[3,3],[0,1],[0,0]]

# output:
[[1,6],[8,10],[15,18]]
[[1,5]]
[[1,4]]
[[1,10]]
[[0,1],[3,3]]

 

先对输入排序，使用C++比较器格式：

    vector> merge(vector>& intervals) {
        if(intervals.empty())
            return {};
        vector>res;
        //keypoint
        sort(begin(intervals), end(intervals),
             [](vector a, vector b){return a[0]

Python:

    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        if len(intervals)==0: return []
        intervals=sorted(intervals,key=lambda x : x[0])
        res=[intervals[0]]
        for n in intervals[1:]:
            if n[0] <= res[-1][1]:
                res[-1][1]=max(n[1],res[-1][1])
            else:
                res.append(n)
        return res