【leetcode】Insert Interval

前言：这一题的难度为hard，题目描述比较简单，思想也不复杂，陷阱就在于很多情况都要考虑，细节问题要多注意。

除了题目中给出的两种情况，下面是我出错的测试用例

第一类测试用例：
原数组：[[3,5],[12,15]] 插入：[6,6]
期望结果：[[3,5],[6,6],[12,15]]

第二类测试用例：
原数组：[[0,5],[9,12]]，插入：[7, 16]
期望结果：[[0,5],[7,16]]

第三类测试用例：
原数组：[[1,5]]，插入：[0,3]
期望结果：[[0,5]]

第四类测试用例：
原数组：[[1,5]]，插入：[6,8]
期望结果：[[1,5],[6,8]]

最后成功的代码为：

{    if(intervals.size() == 0 || newInterval.start > intervals.rbegin() -> end)    {        intervals.push_back(newInterval);        return intervals;    }     if(newInterval.end < intervals.begin() -> start)    {        intervals.insert(intervals.begin(), newInterval);        return intervals;     }    vector<Interval>::iterator left = intervals.begin(), right = intervals.end();    vector<Interval>::iterator mid;    while(left < right)    {        mid = left + (right - left)/2;        //cout << mid -> start << " " << mid -> end << endl;        if(newInterval.start >= mid -> start && newInterval.start <= mid -> end)            break;        else if(newInterval.start < mid -> start)            right = mid;        else if(newInterval.start > mid -> end)            left = mid + 1;    }     vector<Interval>::iterator sp = mid;    if(newInterval.start > mid->end)        sp += 1;    cout << sp -> start << " " << sp -> end << endl;    left = intervals.begin();    right = intervals.end();    while(left < right)    {        mid = left + (right - left)/2;        if(newInterval.end >= mid -> start && newInterval.end <= mid -> end)            break;        else if(newInterval.end < mid -> start)            right = mid;        else if(newInterval.end > mid -> end)            left = mid + 1;    }        vector<Interval>::iterator ep = mid;    if(newInterval.end < mid->start)        ep -= 1;    cout << ep -> start << " " << ep -> end << endl;    if(ep < sp)    {        intervals.insert(sp, newInterval);    }    else    {        ep -> end = ep -> end > newInterval.end ? ep -> end : newInterval.end;        sp -> start = sp -> start < newInterval.start ? sp -> start : newInterval.start;        if(sp < ep)        {            sp -> end = ep -> end;            intervals.erase(sp + 1, ep + 1);        }    }    return intervals;}