[LeetCode] Merge Intervals
2014-01-04 09:28
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问题:
Given a collection of intervals, merge all overlapping intervals.
For example,
Given
return
分析:
不难的一道题,但需要注意C++中的comparator的用法。参见这里。
代码:
bool comp(const Interval &v1, const Interval &v2) {
return v1.start < v2.start;
}
class Solution {
public:
vector<Interval> merge(vector<Interval> &intervals) {
vector<Interval> result;
if (intervals.empty())
return result;
std::sort(intervals.begin(), intervals.end(), comp);
result.push_back(intervals[0]);
for (int i = 1; i < intervals.size(); i ++) {
if (intervals[i].start <= result.back().end)
result.back().end = max(result.back().end, intervals[i].end);
else
result.push_back(intervals[i]);
}
return result;
}
};
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].
分析:
不难的一道题,但需要注意C++中的comparator的用法。参见这里。
代码:
bool comp(const Interval &v1, const Interval &v2) {
return v1.start < v2.start;
}
class Solution {
public:
vector<Interval> merge(vector<Interval> &intervals) {
vector<Interval> result;
if (intervals.empty())
return result;
std::sort(intervals.begin(), intervals.end(), comp);
result.push_back(intervals[0]);
for (int i = 1; i < intervals.size(); i ++) {
if (intervals[i].start <= result.back().end)
result.back().end = max(result.back().end, intervals[i].end);
else
result.push_back(intervals[i]);
}
return result;
}
};
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