【数据结构基础_数组】Leetcode 56.合并区间

原题链接：Leetcode 56. Merge Intervals

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

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

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

• 1 <= intervals.length <= 10⁴
• intervals[i].length == 2
• 0 <= starti <= endi <= 10⁴

---

方法一：排序

思路：

首先对所有区间按左节点升序排序
然后遍历每个区间：

• 当前节点的左节点处于上个区间中时，用当前节点的右节点更新上个区间的右节点
• 否则说明当前区间和上个区间是不相交的，当前区间直接加入结果集

C++代码：

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        // 提前返回
        if (intervals.size() == 0) {
            return {};
        }

        // 按左端点升序排列
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> ans;

        // 遍历每个区间
        for (int i = 0; i < intervals.size(); ++i) {
            // lr左右端点
            int l = intervals[i][0], r = intervals[i][1];

            // 分隔的区间 加入结果
            if (!ans.size() || ans.back()[1] < l) {
                ans.push_back({l, r});
            }
            // 区间合并
            else {
                ans.back()[1] = max(ans.back()[1], r);
            }
        }
        return ans;
    }
};


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复杂度分析：

• 时间复杂度：O(nlogn)，排序占了大头
• 空间复杂度：O(logn)，排序用掉的空间