【Leetcode】Insert Interval

题目：

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

分析：

其实这题跟之前对区间的合并非常相似，可以当作先给整体区间进行元素的添加，然后再进行区间的合并。

Java版本：

/** * Definition for an interval. * public class Interval { *     int start; *     int end; *     Interval() { start = 0; end = 0; } *     Interval(int s, int e) { start = s; end = e; } * } */class Solution {    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {        int n = intervals.size() + 1;        int[] starts = new int[n];        int[] ends = new int[n];        intervals.add(newInterval);        for(int i = 0; i < n; i++){            starts[i] = intervals.get(i).start;            ends[i] = intervals.get(i).end;                    }        Arrays.sort(starts);        Arrays.sort(ends);        List<Interval> res = new ArrayList<Interval>();        for( int i=0,  j = 0; i < n; i++){            if( i == n - 1 || starts[i+1] > ends[i]  ){                res.add(new Interval(starts[j], ends[i]));                j = i + 1;            }        }        return res;    }}

Python版本：

# Definition for an interval.# class Interval:#     def __init__(self, s=0, e=0):#         self.start = s#         self.end = eclass Solution:    def insert(self, intervals, newInterval):        """        :type intervals: List[Interval]        :type newInterval: Interval        :rtype: List[Interval]        """        if len(intervals) == 0 : return [];        intervals.append(newInterval);        intervals = sorted(intervals, key = lambda x : x.start)        res = [intervals[0]]        for n in intervals[1:] :            if n.start <= res[-1].end : res[-1].end = max(n.end,res[-1].end)            else: res.append(n)        return res;

下面我们换一种思路，采用C++中STL的一些高阶函数：
C++版本：

/** * 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<Interval> insert(vector<Interval>& intervals, Interval newInterval) {        auto compare = [] (const Interval &intv1, const Interval &intv2)                          { return intv1.end < intv2.start; };        auto range = equal_range(intervals.begin(), intervals.end(), newInterval, compare);        auto itr1 = range.first, itr2 = range.second;        if (itr1 == itr2) {            intervals.insert(itr1, newInterval);        } else {            itr2--;            itr2->start = min(newInterval.start, itr1->start);            itr2->end = max(newInterval.end, itr2->end);            intervals.erase(itr1, itr2);        }        return intervals;    }};