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.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

AC代码

bool cmp(const vector<int>& a, const vector<int>& b) { return a[0] < b[0]; }

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(), cmp);
        vector<vector<int>> ans;
        for (auto interval : intervals) {
            if (ans.empty()) {
                vector<int> t{interval[0], interval[1]};
                ans.push_back(t);
            }
            else {
                if (interval[0] <= ans[ans.size() - 1][1])
                    ans[ans.size() - 1][1] = max(ans[ans.size() - 1][1], interval[1]);
                else {
                    vector<int> t{interval[0], interval[1]};
                    ans.push_back(t);
                }
            }
        }
        return ans;
    }
};

总结

将所有间隔按照左端点排序，逐一比较右端点，若下个组间隔的右端点小于上一组的，那么就有重叠，比较两个右端点，取其大者。没有重叠，就往结果vector<vector<int>>中添加新元素