题解 | #合并区间#

/**
 * 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) {
        // 思路：
        // (1) 将intervals首个元素放到ans中
        // (2) 遍历intervals的剩余元素elem，如果ans.back()大于elem.start范围内，更新back的值
        // (3) 如果不在，那么就将其保存到ans中
        if(intervals.empty()){
            return {};
        }
        sort(intervals.begin(), intervals.end(), [&](Interval& a, Interval& b){
            return a.start == b.start ? a.end < b.end : a.start < b.start;
        });
        vector<Interval> ans;
        int size = intervals.size();
        ans.push_back(intervals[0]);
        for(int i = 1; i < size; ++i){
            int end = ans.back().end;
            Interval elem= intervals[i];
            if( end >= elem.start){
                ans.back().end = max(end, elem.end);
            }else{
                ans.push_back(elem);
            }
        }
        return ans;
    }
};