区间调度之区间`合并`

问题:给出若干区间,区间之间可能有交集。输出合并之后的区间。

例题

1.Merge Intervals

https://leetcode-cn.com/problems/merge-intervals/

思路

按照区间的左端点,右端点从小道大排序。遍历每一个区间,若当前区间curr.start <= last.end,证明区间有可能有交集,需要更新上一个区间last.end的值。若curr.start > last.end 证明两个区间不可能有交集。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct Record {
    int left, right;
};
bool cmp(Record a, Record b) {
    if (a.left != b.left) {
        return a.left < b.left;
    }
    else {
        return a.right < b.right;
    }
}
class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        int n = intervals.size();
        if (n == 0) {
            return *(new vector<vector<int>>);
        }
        Record* rec = new Record[n];
        for (int i = 0; i < n; i++) {
            rec[i].left = intervals[i][0];
            rec[i].right = intervals[i][1];
        }
        sort(rec, rec + n, cmp);
        Record* ans = new Record[n];
        int k = 0;
        ans[k].left = rec[k].left;
        ans[k].right = rec[k].right;
        for (int curr = 1; curr < n; curr++) {
            if (ans[k].right >= rec[curr].left&&rec[curr].right >ans[k].right) {
                ans[k].right = rec[curr].right;
            }
            else if (ans[k].right < rec[curr].left) {
                ans[++k].left = rec[curr].left;
                ans[k].right = rec[curr].right;
            }
        }
        vector < vector<int>>* ans_vc = new vector < vector<int>>;
        for (int i = 0; i <= k; i++) {
            vector<int>* temp = new vector<int>;
            temp->push_back(ans[i].left);
            temp->push_back(ans[i].right);
            ans_vc->push_back(*temp);
        }
        return *ans_vc;
    }
};

2.剩下的树

https://www.nowcoder.com/practice/f5787c69f5cf41499ba4706bc93700a2?tpId=40&tqId=21356&tPage=2&rp=1&ru=%2Fta%2Fkaoyan&qru=%2Fta%2Fkaoyan%2Fquestion-ranking

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct node
{
	int left, right;
};
bool cmp(node a, node b) {
	if (a.left != b.left) {
		return a.left < b.left;
	}
	else {
		return a.right < b.right;
	}
}
int main()
{
	//freopen("in.text", "r", stdin);
	int n,m;
	while (cin>>n>>m)
	{
		node* rec = new node[m];
		for (int i = 0; i < m; i++)
		{
			cin >> rec[i].left >> rec[i].right;
		}
		sort(rec, rec + m, cmp);
		vector<node> vc;
		node temp;
		temp.left = rec[0].left;
		temp.right = rec[0].right;
		vc.push_back(temp);
		for (int curr = 1; curr < m; curr++)
		{
			int last = vc.size() - 1;
			if (vc[last].right >= rec[curr].left && vc[last].right < rec[curr].right) {
				vc[last].right = rec[curr].right;
			}
			else if(rec[curr].left>vc[last].right){
				temp.left = rec[curr].left;
				temp.right = rec[curr].right;
				vc.push_back(temp);
			}
		}

		
		int ans = (n + 1);
		for (int i = 0; i < vc.size(); i++)
		{
			//cout << "[" << vc[i].left << "," << vc[i].right << "]" << endl;
			ans -= (vc[i].right - vc[i].left + 1);
		}
		cout << ans << endl;
	}
	//fclose(stdin);
}

原文地址:https://www.cnblogs.com/custoyth/p/12720281.html