力扣算法题—057插入区间

给出一个无重叠的 ，按照区间起始端点排序的区间列表。

在列表中插入一个新的区间，你需要确保列表中的区间仍然有序且不重叠（如果有必要的话，可以合并区间）。

示例 1:

输入: intervals = [[1,3],[6,9]], newInterval = [2,5]
输出: [[1,5],[6,9]]

示例 2:

输入: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
输出: [[1,2],[3,10],[12,16]]
解释: 这是因为新的区间 [4,8] 与 [3,5],[6,7],[8,10] 重叠。

#include "000库函数.h"



struct Interval {
    int start;
    int end;
    Interval() : start(0), end(0) {}
    Interval(int s, int e) : start(s), end(e) {}
};
 
//与上题合并区间没什么区别
//只不过将区间加入进去而已

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        if (intervals.empty())return { newInterval };
        vector<Interval>res;
        vector<int>s, e;
        for (int i = 0; i < intervals.size(); ++i) {
            s.push_back(intervals[i].start);
            e.push_back(intervals[i].end);
        }
        s.push_back(newInterval.start);
        e.push_back(newInterval.end);
        sort(s.begin(), s.end());//头排序
        sort(e.begin(), e.end());//尾排序
        Interval p;
        p.start = s[0];
        for (int i = 1; i < s.size(); ++i) {
            if ((s[i] - e[i - 1]) < 1)continue;//比较大小
            p.end = e[i - 1];
            res.push_back(p);
            p.start = s[i];
        }
        p.end = e[s.size() - 1];
        res.push_back(p);
        return res;
    }
};


//题目中确保原来的区间是按起始端且无重叠，则复杂度较上题更小
class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        vector<Interval> res;
        int n = intervals.size(), cur = 0;
        while (cur < n && intervals[cur].end < newInterval.start) {
            res.push_back(intervals[cur++]);
        }
        while (cur < n && intervals[cur].start <= newInterval.end) {
            newInterval.start = min(newInterval.start, intervals[cur].start);
            newInterval.end = max(newInterval.end, intervals[cur].end);
            ++cur;
        }
        res.push_back(newInterval);
        while (cur < n) {
            res.push_back(intervals[cur++]);
        }
        return res;
    }
};






void T057() {
    Solution s;
    vector<Interval>v;
    Interval newInterval;
    vector<int>nums;
    Interval p;
    nums = { 1,4,8,10 };
    for (int i = 0; i < nums.size(); i += 2) {
        p.start = nums[i];
        p.end = nums[i + 1];
        v.push_back(p);
    }
    newInterval.start = 5;
    newInterval.end = 7;
    v = s.insert(v, newInterval);
    for (auto a : v)
        cout << a.start << ", " << a.end << ";  ";
    cout << endl;

    v.clear();
    nums = { 1,4,8,10 };
    for (int i = 0; i < nums.size(); i += 2) {
        p.start = nums[i];
        p.end = nums[i + 1];
        v.push_back(p);
    }
    newInterval.start = 3;
    newInterval.end = 5;
    v = s.insert(v, newInterval);
    for (auto a : v)
        cout << a.start << ", " << a.end << ";  ";
    cout << endl;

    v.clear();
    nums = { 1,4,8,10 };
    for (int i = 0; i < nums.size(); i += 2) {
        p.start = nums[i];
        p.end = nums[i + 1];
        v.push_back(p);
    }
    newInterval.start = 0;
    newInterval.end = 7;
    v = s.insert(v, newInterval);
    for (auto a : v)
        cout << a.start << ", " << a.end << ";  ";
    cout << endl;

    v.clear();
    nums = { 1,4,8,10 };
    for (int i = 0; i < nums.size(); i += 2) {
        p.start = nums[i];
        p.end = nums[i + 1];
        v.push_back(p);
    }
    newInterval.start = 2;
    newInterval.end = 9;
    v = s.insert(v, newInterval);
    for (auto a : v)
        cout << a.start << ", " << a.end << ";  ";
    cout << endl;

    
}