[LeetCode] Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considerred overlapping.
混合包含的区间
思路:
0. 需要将给定数组按照start的升序排序,当start相同时,按end升序排序。
1. 需要混合的区间包含如下情况
  1. 如果start相同,若后一个元素的end大于前一个元素end,则融合
  2. 如果start不同,若后一个元素的end大于前一个元素end,则融合
2. 不需要融合的情况
  1. 前一个元素end小于后一个元素的start
参考代码如下:
/**
 * 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:
    static bool cmp(const Interval& a, const Interval& b)
    {
        return a.start < b.start || (a.start == b.start && a.end < b.end);
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), cmp);
        vector<Interval> res;
        for (int i = 0; i < intervals.size(); ++i)
        {
            if (i == 0 || intervals[i].start > res.back().end)
                res.push_back(intervals[i]);
            else if (intervals[i].end > res.back().end)
                res.back().end = intervals[i].end;
        }
        return res;
    }
};

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转载自www.cnblogs.com/immjc/p/9066534.html