LintCode 30: Insert Interval

Description
Given a non-overlapping interval list which is sorted by start point.

Insert a new interval into it, make sure the list is still in order and non-overlapping (merge intervals if necessary).

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Example
Insert (2, 5) into [(1,2), (5,9)], we get [(1,9)].

Insert (3, 4) into [(1,2), (5,9)], we get [(1,2), (3,4), (5,9)].

解法：
分3种情况分别处理。

/**
 * Definition of Interval:
 * classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this->start = start;
 *         this->end = end;
 *     }
 * 
 * }
 */

class Solution {
public:
    /**
     * @param intervals: Sorted interval list.
     * @param newInterval: new interval.
     * @return: A new interval list.
     */
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        vector<Interval> res;
        int n = intervals.size(), cur = 0;
        for (int i = 0; i < n; ++i) {
            if (intervals[i].end < newInterval.start) {
                res.push_back(intervals[i]);
                ++cur;
            } else if (intervals[i].start > newInterval.end) {
                res.push_back(intervals[i]);
            } else {
                newInterval.start = min(newInterval.start, intervals[i].start);
                newInterval.end = max(newInterval.end, intervals[i].end);
            }
        }
        res.insert(res.begin() + cur, newInterval);
        return res;
    }
};

解法2：这个解法可能还好点，因为最后不用在vector中间插入。
注意：需要插入newInterval的情况只有两种。

1. interval.start > newInterval.end，这个时候newInterval已经处理完所有的overlap，需要插入在interval的前面。
2. 全部处理完了，newInterval放在最后。

class Solution {
public:
    /**
     * @param intervals: Sorted interval list.
     * @param newInterval: new interval.
     * @return: A new interval list.
     */
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> res;
        bool inSerted = false;
        for (auto interval : intervals) {
            if (interval.start > newInterval.end) {
                if (!inSerted) { 
                    res.push_back(newInterval);
                    inSerted = true;
                }
                res.push_back(interval);
            } else if (interval.end < newInterval.start) {
                res.push_back(interval);
            } else {
                newInterval.start = min(interval.start, newInterval.start);
                newInterval.end = max(interval.end, newInterval.end);
            }
        }
        if (!inSerted) {
            res.push_back(newInterval);
        }
        return res;
    }
};

解法3：跟解法2差不多，更简单。
把结果的result分成left和right两部分。
left处理完后把newInterval push_back到left，
没有冲突的都放到right。
最后left和right再合并起来。
下次再做。