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: 1Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [ [1,2], [1,2], [1,2] ]Output: 2Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [ [1,2], [2,3] ]Output: 0Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
这道题给了我们一堆区间，让求需要至少移除多少个区间才能使剩下的区间没有重叠，那么首先要给区间排序，根据每个区间的 start 来做升序排序，然后开始要查找重叠区间，判断方法是看如果前一个区间的 end 大于后一个区间的 start，那么一定是重复区间，此时结果 res 自增1，我们需要删除一个，那么此时究竟该删哪一个呢，为了保证总体去掉的区间数最小，我们去掉那个 end 值较大的区间，而在代码中，我们并没有真正的删掉某一个区间，而是用一个变量 last 指向上一个需要比较的区间，我们将 last 指向 end 值较小的那个区间；如果两个区间没有重叠，那么此时 last 指向当前区间，继续进行下一次遍历，参见代码如下：
解法一：
class Solution {public: int eraseOverlapIntervals(vector >& intervals) { int res = 0, n = intervals.size(), last = 0; sort(intervals.begin(), intervals.end()); for (int i = 1; i < n; ++i) { if (intervals[i][0] < intervals[last][1]) { ++res; if (intervals[i][1] < intervals[last][1]) last = i; } else { last = i; } } return res; }};
我们也可以对上面代码进行简化，主要利用三元操作符来代替 if 从句，参见代码如下：
解法二：
class Solution {public: int eraseOverlapIntervals(vector >& intervals) { if (intervals.empty()) return 0; sort(intervals.begin(), intervals.end()); int res = 0, n = intervals.size(), endLast = intervals[0][1]; for (int i = 1; i < n; ++i) { int t = endLast > intervals[i][0] ? 1 : 0; endLast = t == 1 ? min(endLast, intervals[i][1]) : intervals[i][1]; res += t; } return res; }};
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