56. Merge Intervals

Description：
Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Analysis:
题目没说有序，需要先排序。
有交集的时候，才合并。

Solution1:
自己想的

class Solution {
public:
    static bool compare(vector<int>& q1, vector<int>& q2) {
        if (q1[0] < q2[0])
            return true;
        else
            return false;
    }
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>> ret;
        if (intervals.size() == 0)
            return ret;
        sort(intervals.begin(), intervals.end(), compare);
        vector<int> cur;
        bool overlap = false;
        int low = 0, high = 0;
        int i = 0;
        for (; i < intervals.size();) {
            if (overlap) {
                if (intervals[i][0] > cur[1] || intervals[i][1] < cur[0]) {
                    ret.push_back(cur);
                    cur.clear();
                    overlap = false;
                } else {
                    low = min(intervals[i][0], cur[0]);
                    high = max(intervals[i][1], cur[1]);
                    cur.clear();
                    cur.push_back(low);
                    cur.push_back(high);
                    ++ i;
                }
            } else {
                if (i == intervals.size() - 1)
                    break;
                if (intervals[i][0] > intervals[i + 1][1] || intervals[i][1] < intervals[i + 1][0]) {
                    ret.push_back(intervals[i]);
                    ++ i;
                    continue;
                }
                overlap = true;
                low = min(intervals[i][0], intervals[i + 1][0]);
                high = max(intervals[i][1], intervals[i + 1][1]);
                cur.push_back(low);
                cur.push_back(high);
                ++ i;
                ++ i;
            }
        }
        if (overlap)
            ret.push_back(cur);
        if (i == intervals.size() - 1)
            ret.push_back(intervals[i]);
        return ret;
    }
};

Solution2:
参考：双指针法
1.先排序
2.左指针指向可能存在合并的起始区间，右指针开始向后遍历查找
3.右指针发现可以合并，记录当前右端点，继续向后
4.直到右指针走不动，将当前左端点、右端点加入返回值
5.右指针赋给左指针

vector<vector<int>> merge(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> ans;
        for (int i = 0; i < intervals.size();) {
            int t = intervals[i][1];
            int j = i + 1;
            while (j < intervals.size() && intervals[j][0] <= t) {
                t = max(t, intervals[j][1]);
                j++;
            }
            ans.push_back({ intervals[i][0], t });
            i = j;
        }
        return ans;
    }