435. 无重叠区间
1 class Solution 2 { 3 public: 4 int eraseOverlapIntervals(vector<vector<int>>& intervals) 5 { 6 if(intervals.empty() || intervals[0].empty()) return 0; 7 sort(intervals.begin(),intervals.end()); 8 9 int begin = intervals[0][0],end = intervals[0][1]; 10 int count = 0; 11 for(int i = 1;i < intervals.size();i ++) 12 { 13 if(intervals[i][1] < end) 14 { 15 begin = intervals[i][0]; 16 end = intervals[i][1]; 17 count ++; 18 } 19 else if(intervals[i][1] >= end && intervals[i][0] < end) 20 { 21 begin = intervals[i][0]; 22 count ++; 23 } //区间[1,2]和[2,3]的边界相互“接触”,但没有相互重叠 24 else if(intervals[i][0] >= end) 25 { 26 begin = intervals[i][0]; 27 end = intervals[i][1]; 28 } 29 } 30 return count; 31 } 32 };
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