leetcode Merge Intervals
2017-09-22 21:26
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Merge Intervals
题目详情:
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].
解题方法:
首先将按照每组数第一个数的的大小给这些·组数排序,之后从第一组开始到最后一组,若能合并则合并,不能合并则在向后 移一组。详细代码:
/*** 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> merge(vector<Interval>& intervals) {
vector<Interval> result;
if (intervals.size() <= 1) {
return intervals;
}
for (int j = 0; j < intervals.size()-1; j++)
for (int i = 0; i < intervals.size()-1-j; i++) {
if (intervals[i].start > intervals[i+1].start) {
Interval temp(intervals[i].start, intervals[i].end);
intervals[i].start = intervals[i+1].start;
intervals[i].end = intervals[i+1].end;
intervals[i+1].start = temp.start;
intervals[i+1].end = temp.end;
}
}
int j = 0;
result.push_back(intervals[j]);
for (int i = 1; i < intervals.size(); i++) {
if (result[j].end >= intervals[i].start) {
result[j].end = (result[j].end < intervals[i].end ? intervals[i].end:result[j].end);
} else {
result.push_back(intervals[i]);
j++;
}
}
return result;
}
};
复杂度分析:
冒泡排序,复杂度为O(n^2)相关文章推荐
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