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60. Insert Interval && Merge Intervals
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].
思路： 因为区间按 start 升序，且无重叠。所以插入区间和每一个元素分三种情况考虑。在左边，在右边（此两种情况直接拿区间出来）或者交叉（则更新插入区间范围）。 利用变量 out 判断新的区间是否已经放入。
/** * 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) { vector<Interval> vec; bool out = true; for(size_t i = 0; i < intervals.size(); ++i) { if(intervals[i].end < newInterval.start) { vec.push_back(intervals[i]); } else if(intervals[i].start > newInterval.end) { if(out) { vec.push_back(newInterval); out = false;} vec.push_back(intervals[i]); } else { newInterval.start = min(newInterval.start, intervals[i].start); newInterval.end = max(newInterval.end, intervals[i].end); } } if(out) vec.push_back(newInterval); return vec; }};
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].
思路： 先按 start 排序。然后，判断当前区间和前一区间是否重叠。若没重叠，则放入；若重叠，则更新前一区间 end, 舍弃当前区间。
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */bool cmp(Interval a, Interval b) { return a.start < b.start;}class Solution {public: vector<Interval> merge(vector<Interval> &intervals) { sort(intervals.begin(), intervals.end(), cmp); vector<Interval> vec; for(size_t i = 0; i < intervals.size(); ++i) { if(vec.empty()) vec.push_back(intervals[i]); else if(intervals[i].start <= vec.back().end) vec.back().end = max(intervals[i].end, vec.back().end); else vec.push_back(intervals[i]); } return vec; }};
60. Insert Interval && Merge Intervals