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[LeetCode] 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].
先把要插入的区间放到最后，再排序，然后就跟上题一一模一样了。
1 /** 2 * Definition for an interval. 3 * struct Interval { 4 * int start; 5 * int end; 6 * Interval() : start(0), end(0) {} 7 * Interval(int s, int e) : start(s), end(e) {} 8 * }; 9 */10 11 bool cmp(const Interval &a, const Interval &b) {12 return a.start < b.start;13 }14 15 class Solution {16 public:17 vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {18 int pos = 0, cnt = 0;19 intervals.push_back(newInterval);20 sort(intervals.begin(), intervals.end(), cmp);21 for (int i = 1; i < intervals.size(); ++i) {22 if (intervals[pos].end >= intervals[i].start) {23 ++cnt;24 if (intervals[pos].end < intervals[i].end) {25 intervals[pos].end = intervals[i].end;26 }27 } else {28 ++pos;29 intervals[pos].start = intervals[i].start;30 intervals[pos].end = intervals[i].end;31 }32 }33 intervals.resize(intervals.size()-cnt);34 return intervals;35 }36 };
[LeetCode] Insert Interval
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