# 56. Merge Intervals

## 1.問題

* 給予一個區間組成的集合, merge所有重疊的區間

![](/files/-LJ2LHGylop-p1EPZj9w)

## 2.想法

* 提問: interval間是否已排序?
* parameter
  * Interval list
* test input
* 說明想法

  * 必須先排序Interval: overwrite sort operation

  ```
  //define sort operation
  struct IntervalKey {
          inline bool operator() (const Interval &pLHS, const Interval &pRHS) {
              return pLHS.start < pRHS.start;
          }
      };

  //sort vector
  sort(intervals.begin(), intervals.end(), IntervalKey());
  ```

  * 比較Interval\[i]與interval\[cnt]:
    * 如果有重疊, 則讓interval\[cnt].end = interval\[i].end
    * 不然則將cnt++, 讓interval\[i]加入新的interval\[cnt]
* 測試計算複雜度

## 3.程式碼

```
/**
 * 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:
    struct intervalKey {
        inline bool operator()(const Interval& in1, const Interval& in2) {
            return in1.start < in2.start;
        }
    };
    
    vector<Interval> merge(vector<Interval>& intervals) {
        vector<Interval> res;
        if (intervals.size() == 0) {
            return res;
        }
        sort(intervals.begin(), intervals.end(), intervalKey());
        int cnt = 0, n = intervals.size();
        res.push_back(intervals[0]);
        for (int i = 1; i < n; i++) {
            if (intervals[i].start <= res[cnt].end) {
                res[cnt].end = max(intervals[i].end, res[cnt].end);
            } else {
                cnt++;
                res.push_back(intervals[i]);
            }
        }
        
        return res;
    }
};
```

## 4.Performance

![](/files/-LJ2MVhW0o2cqcDD8YoY)


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