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

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