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1. Merge
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].
/** * 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: static bool compare(Interval a, Interval b) { if(a.start != b.start) return a.start < b.start; return a.end < b.end; } vector<Interval> merge(vector<Interval>& intervals) { sort(intervals.begin(), intervals.end(), compare); vector<Interval> v; int n = intervals.size(), i; if(n<1) return v; for(i = 0; i < n; i++) { if(v.size() && intervals[i].start <= v[v.size()-1].end) v[v.size()-1].end = max(intervals[i].end, v[v.size()-1].end); else v.push_back(intervals[i]); } return v; } };
2. Insert
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].
(1) 用上面的merge方法
/** * 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: static bool compare(Interval a, Interval b) { if(a.start != b.start) return a.start < b.start; return a.end < b.end; } vector<Interval> merge(vector<Interval>& intervals) { sort(intervals.begin(), intervals.end(), compare); vector<Interval> v; int n = intervals.size(), i; if(n<1) return v; for(i = 0; i < n; i++) { if(v.size() && intervals[i].start <= v[v.size()-1].end) v[v.size()-1].end = max(intervals[i].end, v[v.size()-1].end); else v.push_back(intervals[i]); } return v; } vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) { intervals.push_back(newInterval); return merge(intervals); } };
(2)
/** * 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 n = intervals.size(), l, i; vector<Interval> v; for(i = 0; i < n; i++) { if(newInterval.start > intervals[i].end) v.push_back(intervals[i]); else break; } if(i >= n) { v.push_back(newInterval); return v; } l = v.size(); v.push_back(intervals[i]); if(newInterval.start < v[l].start) v[l].start = newInterval.start; while(i < n && newInterval.end > intervals[i].end) i++; if(i<n && newInterval.end >= intervals[i].start) v[l].end = intervals[i++].end; else v[l].end = newInterval.end; while(i<n) v.push_back(intervals[i++]); return v; } };
56. Merge Intervals 57. Insert Interval *HARD*
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原文地址：http://www.cnblogs.com/argenbarbie/p/5266607.html
