leetcode 57.Insert Interval
题意:
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]
.
刚开始遇到这个问题,我看了下hint,以为要用到binary tree,但是感觉上应该是挺简单的。后来发现却是是挺简单的
主要份两种情况:
1.newintervals 与原来的intervals 有交集,那么就需要合并
2.newintervals与原来的intervals 无交集,那么就不需要合并,直接按照大小排列即可
然后直接开一个新的数组,用于返回值,这样利用空间节省时间。
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */ class Solution { public: vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { vector<Interval> ret; if(intervals.empty()) { ret.push_back(newInterval); return ret; } int i = 0; while(i < intervals.size()) { //no overlapping if(newInterval.end < intervals[i].start) { ret.push_back(newInterval); while(i < intervals.size()) { ret.push_back(intervals[i]); i ++; } return ret; } else if(newInterval.start > intervals[i].end) ret.push_back(intervals[i]); //overlapping else { newInterval.start = min(newInterval.start, intervals[i].start); newInterval.end = max(newInterval.end, intervals[i].end); } i ++; } ret.push_back(newInterval); return ret; } };