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题目描述:
给出一个区间的集合,请合并所有重叠的区间。
示例 1:
输入: [[1,3],[2,6],[8,10],[15,18]] 输出: [[1,6],[8,10],[15,18]] 解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].
示例 2:
输入: [[1,4],[4,5]] 输出: [[1,5]] 解释: 区间 [1,4] 和 [4,5] 可被视为重叠区间。
解题思路:首先将区间进行一次排序,之后一次遍历就行,遍历到的区间和当前res的末区间相比较,需要更新就将res弹出末尾。。继续比较,1.若比到最后res为空,直接将结果放入,跳出循环 2.若不是重叠区间,则放入,跳出循环
C++解题:
class Solution { public: vector<vector<int>> merge(vector<vector<int>>& intervals) { sort(intervals.begin(),intervals.end()); vector<vector<int>> res; for (int i = 0; i < intervals.size(); i++) { if(!res.size()){ res.push_back(intervals[i]); }else{ vector<int> inte_i = intervals[i]; while(1){ vector<int> tmp = res.back(); if(inte_i[0] <= tmp[1])//更新区间 { tmp[1] = tmp[1] > inte_i[1] ? tmp[1]:inte_i[1]; res.pop_back(); inte_i = tmp; if(!res.size()){ res.push_back(tmp); break; } } else //直接加入新区间 { res.push_back(inte_i); break; } } } } return res; } };
The text was updated successfully, but these errors were encountered:
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题目描述:
给出一个区间的集合,请合并所有重叠的区间。
示例 1:
示例 2:
解题思路:首先将区间进行一次排序,之后一次遍历就行,遍历到的区间和当前res的末区间相比较,需要更新就将res弹出末尾。。继续比较,1.若比到最后res为空,直接将结果放入,跳出循环 2.若不是重叠区间,则放入,跳出循环
C++解题:
The text was updated successfully, but these errors were encountered: