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Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

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:

• 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.

题意：给出一个区间集合，移除最少数量的区间，使得剩余区间不重合。

解法 贪心

本题等同于——最多保留多少个区间，让它们之间互相不重叠。关键是按照区间右端点从小到大来进行排序。具体代码如下：

class Solution {
   public:    int eraseOverlapIntervals(vector
   
    
     >& intervals) {
           if (intervals.empty()) return 0;         sort(intervals.begin(), intervals.end(), [&](const vector
     
       &a, const vector
      
        &b) {
               return a[1] != b[1] ? a[1] < b[1] : a[0] < b[0];        });        int nonOverlapping = 0, n = intervals.size(), end = INT_MIN;        for (int i = 0; i < n; ++i) {
               if (intervals[i][0] >= end) {
                   ++nonOverlapping;                end = intervals[i][1];             }         }        return n - nonOverlapping;    }};
      
     
    
   

运行效率如下：

执行用时：32 ms, 在所有 C++ 提交中击败了69.63% 的用户内存消耗：9.3 MB, 在所有 C++ 提交中击败了78.73% 的用户

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