LeetCode OJ Insert Interval

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Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

class Solution {public:    void include(vector<Interval> &intervals, int target, int & l, int & r, bool & isIncluded) {  // 判断target是否在intervals中，如果在，l、r都为区间坐标，否则为左右区间坐标        for (int i = 0; i < intervals.size(); i++) {            if (target < intervals[i].start) {                l = i - 1;                r = i;                isIncluded = false;                return;            } else if (intervals[i].start <= target && target <= intervals[i].end) {                l = r = i;                isIncluded = true;                return;            }        }        l = intervals.size() - 1;        r = intervals.size();        isIncluded = false;    }    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {        vector<Interval> ans;        int sl, sr, el, er;        bool si, ei;        include(intervals, newInterval.start, sl, sr, si);        include(intervals, newInterval.end, el, er, ei);        if (si && ei) {  // 如果两个都在区间中，只需改变区间上下限            for (int i = 0; i < intervals.size(); i++) {                if (i == sl) {                    Interval newinter(intervals[sl].start, intervals[er].end);                    ans.push_back(newinter);                    i = er;                } else {                    ans.push_back(intervals[i]);                }            }        } else if (si && !ei) {  // 如果仅有一个在区间中            for (int i = 0; i < intervals.size(); i++) {                if (i == sl) {                    Interval newinter(intervals[i].start, newInterval.end);                    ans.push_back(newinter);                    i = el;                } else {                    ans.push_back(intervals[i]);                }            }        } else if (!si && ei) {  // 如果仅有一个在区间中            for (int i = 0; i < intervals.size(); i++) {                if (i == sr) {                    Interval newinter(newInterval.start, intervals[er].end);                    ans.push_back(newinter);                    i = er;                } else {                    ans.push_back(intervals[i]);                }            }        } else {  // 如果两个都不在区间中            if (sr == intervals.size()) {  // 新区间在末尾                ans = intervals;                ans.push_back(Interval(newInterval.start, newInterval.end));                return ans;            }            if (er == 0) {  // 新区间在开头                ans.push_back(Interval(newInterval.start, newInterval.end));                for (int i = 0; i < intervals.size(); i++) ans.push_back(intervals[i]);                return ans;            }            if (sl == el && sr == er) {  // 如果是中间单独形成的一个新区间                for (int i = 0; i < er; i++) ans.push_back(intervals[i]);                ans.push_back(Interval(newInterval.start, newInterval.end));                for (int i = er; i < intervals.size(); i++) ans.push_back(intervals[i]);                return ans;            }            for (int i = 0; i < intervals.size(); i++) {  // 如果包围了原有的区间                if (i == sr) {                    Interval newinter(newInterval.start, newInterval.end);                    ans.push_back(newinter);                    i = el;                } else {                    ans.push_back(intervals[i]);                }            }        }        return ans;    }};

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