Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Note:
- You may assume the interval's end point is always bigger than its start point.
- Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
Example 1:
Input: [ [1,2], [2,3], [3,4], [1,3] ] Output: 1 Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [ [1,2], [1,2], [1,2] ] Output: 2 Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [ [1,2], [2,3] ] Output: 0 Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
解析：利用貪心算法求解，假設在某個區域，有n個區間重疊，則顯然n個區間最後只能留下一個，而根據局部最優，勢必留下end最小的最有利。根據這個思路，事先將數組排序，然後不斷刪去區間即可，時間複雜度爲O(n).
class Solution {
public:
static bool cmp(Interval& a, Interval& b)
{
return a.end < b.end;
}
int eraseOverlapIntervals(vector<Interval>& intervals) {
if(!intervals.size()) return 0;
int num = 0;
sort(intervals.begin(), intervals.end(), cmp);
for(vector<Interval>::iterator i = intervals.begin(); i!=intervals.end(); i++)
while(i != intervals.end()-1 && i->end > (i+1)->start)
{
intervals.erase(i+1);
num ++;
}
return num;
}
};