56/57. Merge Intervals; Insert Interval

56. Merge Intervals

Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

注意：comp函数应声明为static返回类型，否则报错；此外，只有本类会使用，所以声明为private

/** * 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> merge(vector<Interval>& intervals) {        vector<Interval> res;        if( intervals.empty() ) return res;        sort( intervals.begin(), intervals.end(), comp );res.push_back(intervals[0]);        for (int i = 1; i < intervals.size(); ++i) {            if (res.back().end >= intervals[i].start) {                res.back().end = max(res.back().end, intervals[i].end);            }else{                res.push_back(intervals[i]);            }        }        return res;    }private:    static bool comp(const Interval &a, const Interval &b) {        return (a.start<b.start);    }};

168 / 168 test cases passed. Runtime: 26 ms. Your runtime beats 14.31% of cpp submissions.

57. Insert Interval

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].

/** * 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) {        int overlap = 0;        int len = intervals.size();        vector<Interval> res = intervals;        int i=0;        for( ; i<len; ){            if( newInterval.end<intervals[i].start ){                break;            }else{                if( newInterval.start>intervals[i].end ){                }else{                    newInterval.start = min(newInterval.start, res[i].start);                    newInterval.end = max(newInterval.end, res[i].end);                    ++overlap;                }            }            ++i;        }        //delete overlap parts        //if( overlap>0 )        res.erase(res.begin()+i-overlap, res.begin()+i);        res.insert(res.begin()+i-overlap, newInterval);                return res;    }};

152 / 152 test cases passed. Runtime: 16 ms. Your runtime beats 40.28% of cpp submissions.

注：这两个算法的时间性能目前并不好，待优化。