題目描述
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.
給定多組區間，選擇刪除最少的區間，使得剩下的區間無交集。
解題思路
貪心算法，遍歷數組，發現重疊就刪去範圍較大的區間。爲什麼要去掉範圍最大的數組？因爲範圍大，就意味着能與更多的區間產生交集，所以要去掉。解題圖示：
整個步驟如下：
1、對intervals排序，以第一個元素上升排序，第一個相同則按第二個元素上升排序（使用sort排序，自己編寫排序函數com）
2、用tmp表示當前區間，如果與檢查到的區間有交集，tmp變爲兩者之間的較小者，計數器＋1（表示應該刪去較大的那個）
3、無交集，tmp變爲檢查到的那個區間，繼續2（因爲已經排序了，可以保證一旦檢查到第一個與tmp無交集的區間，後面的也就不會有交集了）
代碼如下
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
bool static com(Interval a, Interval b) {
if (a.start < b.start)
return true;
else if (a.start == b.start) {
if (a.end < b.end)
return true;
return false;
}
return false;
}
bool inter(Interval a, Interval b) {
if (a.end > b.start)
return true;
return false;
}
int eraseOverlapIntervals(vector<Interval>& intervals) {
if(intervals.empty())
return 0;
sort(intervals.begin(), intervals.end(), com);
int count = 0;
Interval in = intervals[0];
int j = 1;
for (j; j < intervals.size(); j++) {
if (inter(in, intervals[j])) {
if(in.end > intervals[j].end)
in = intervals[j];
count++;
continue;
}
in = intervals[j];
}
return count;
}
};題外心得：
1、關於sort比較函數的傳入，sort默認是升序排序，要想使得降序，就應該這樣做：
bool com(int a,int b){
return a > b;
}也就是，讓sort認爲 a > b（返回true）時就是 a在b前面。
2、這兩道貪心算法都涉及了排序。
