排序LeetCode 56 Merge Intervals

LeetCode 56

Merge Intervals

• Problem Description:
  合併所有的重疊區間。
  具體的題目信息：
  https://leetcode.com/problems/merge-intervals/description/
• Example:
  
• Solution:
  
  • 解題思路：題目實際考查重疊區間的求法，我們定義一個比較函數將所有的區間按照左區間大小從小到大排序，比較相鄰區間是否有重疊（前者的右區間不小於後者的左區間），如果有重疊，注意更新合併後區間的右區間。
  • 編程實現：

/**
 * 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(Interval p, Interval q) {
        return (p.start<q.start)||(p.start==q.start&&p.end<q.end);
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        if (intervals.size() == 0||intervals.size() == 1) return intervals;
        vector<Interval> res;
        sort(intervals.begin(), intervals.end(), compare);
        int flag = 0, s = 0, e = 0, rear = 0;
        Interval t;
        for (int i = 0; i < intervals.size(); i++) {
            if (flag == 0) {
                if (i < intervals.size()-1) {
                    if (intervals[i].end>=intervals[i+1].start) {
                        flag = 1;
                        s = i;
                        rear = max(intervals[i].end, intervals[i+1].end);
                        if (i+1 == intervals.size()-1) {
                            t = Interval(intervals[s].start, rear);
                            res.push_back(t);
                            break;
                        }
                    } else {
                        res.push_back(intervals[i]);
                    }
                } else {
                    res.push_back(intervals[i]);
                }
            } else {
               while(i < intervals.size()-1&&intervals[i+1].start<=rear) {
                   i++;
                   rear = max(rear, intervals[i].end);
               }
                t = Interval(intervals[s].start, rear);
                res.push_back(t);
                flag = 0;
                if (i == intervals.size()-1) break;
            }
        }
        return res;
    }
};