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].
对若干个区间进行合并，使合并后的区间没有重叠
先对区间按照左边界排序，然后顺序扫描，合并重叠的区间即可。 本文地址
代码1在原区间数组上操作，不使用额外的空间，但是需要删除多余的区间，这样会移动数组元素
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/** * 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 {private: static
bool comp(Interval a, Interval b) { return
a.start < b.start; }public: vector<Interval> merge(vector<Interval> &intervals) { if(intervals.empty())return
intervals; sort(intervals.begin(), intervals.end(), comp); vector<Interval>::iterator it1 = intervals.begin(), it2 = it1 + 1; while(it1 != intervals.end() && it2 != intervals.end()) { if(it2->start <= it1->end) { if(it1->end < it2->end)it1->end = it2->end; it2++; } else { //[it1+1, it2)范围内的区间可以从原数组删除 it1 = intervals.erase(it1+1, it2); it2 = it1 + 1; } } if(it1 != intervals.end()) intervals.erase(it1 + 1, it2); return
intervals; }}; |
代码2用新数组来存储合并后的区间
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/** * 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 {private: static
bool comp(Interval a, Interval b) { return
a.start < b.start; }public: vector<Interval> merge(vector<Interval> &intervals) { if(intervals.empty())return
intervals; sort(intervals.begin(), intervals.end(), comp); vector<Interval> res; res.push_back(intervals[0]); for(int
i = 1; i < intervals.size(); i++) { Interval &p = res.back(); if(intervals[i].start > p.end)res.push_back(intervals[i]); else
if(intervals[i].end > p.end)p.end = intervals[i].end; } return
res; }}; |
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LeetCode:Merge Intervals,布布扣,bubuko.com
原文：http://www.cnblogs.com/TenosDoIt/p/3714681.html