Leetcode | Merge Intervals

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].

这里主要注意的是STL的sort的函数的comparor要写成static函数，而且参数采用值传递，不要用引用。

class Solution {
public:
    static bool comp(Interval m1, Interval m2) {
        return m1.start < m2.start;
    }

    vector<Interval> merge(vector<Interval> &intervals) {
        sort(intervals.begin(), intervals.end(), Solution::comp);
        vector<Interval> ret;
        if (intervals.empty()) return ret;
        Interval pre(intervals[0]);
        for (int i = 0; i < intervals.size(); ++i) {
            if (intervals[i].start <= pre.end) {
                if (intervals[i].end > pre.end) pre.end = intervals[i].end;
            } else {
                ret.push_back(pre);
                pre = intervals[i];
            }
        }
        ret.push_back(pre);
        
        return ret;
    }
};

 这样写。

/**
 * 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 compare(const Interval &i1, const Interval &i2) {
        return i1.start < i2.start;
    }
    vector<Interval> merge(vector<Interval> &intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        vector<Interval> ans;

        int n = intervals.size();
        for (int i = 0; i < n; ++i) {
            if (i == n - 1 || intervals[i].end < intervals[i + 1].start) {
                ans.push_back(intervals[i]);
            } else {
                intervals[i + 1].start = intervals[i].start;
                if (intervals[i].end > intervals[i + 1].end) intervals[i + 1].end = intervals[i].end;
            }
        }
        
        return ans;
    }
};