leetcode 435. Non-overlapping Intervals 消除覆盖区间

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.

这道题题意很简单，就是消除重复区间，就是设置一个前驱区间，嗯嗯，做法很棒的，做法的核心是这样的，当遇到重复区间的时候，每当当前的end小于pre的end的时候，修改pre，

这一道题leetcode 452. Minimum Number of Arrows to Burst Balloons 消除覆盖区间 的做法和本体一样，值得一起学习

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <cmath>using namespace std;/*struct Interval {     int start;     int end;     Interval() : start(0), end(0) {}     Interval(int s, int e) : start(s), end(e) {}};*/bool cmp(Interval a, Interval b){    if (a.start != b.start)        return a.start < b.start;    else        return a.end < b.end;}class Solution {public:    int eraseOverlapIntervals(vector<Interval>& in)     {        sort(in.begin(),in.end(),cmp);        int count = 0 , pre = 0;        for (int now = 1; now < in.size(); now++)        {            if (in[now].start < in[pre].end)            {                count++;                if (in[now].end < in[pre].end)                    pre = now;            }            else                pre = now;        }        return count;    }};