56. Merge Intervals
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.
这道题的解题思路是，先sort从小到大排列一下（按照[x,y]的x排列）。这第一步也是试出来的，本来以为给的数组是排好的，提交后才发现，有的case是顺序打乱的。
例如：
排好顺序以后，我们假设有个区间最小值和最大值。最小值就是第一个括号内左边的值了，最大值初始为第一个括号内右边的值。然后往后逐个判断括号内右边的值是否<=max，如果是，则往右移动，最后记录一个最大的值。得到第一个区间，写到ans中，继续往后历，直到<size()。
AC的答案（C++）：
class Solution {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
vector<vector<int>> ans;
ans.clear();
if(intervals.size()==0)return ans;
vector<int> t;
t.clear();
sort(intervals.begin(),intervals.end());
for(int i=0;i<intervals.size();i++)
{
int x=i;
int y=i+1;
int m=intervals[x][1];
while((y<=intervals.size()-1)&&intervals[y][0]<=m)
{
m=max(m,intervals[y][1]);
y++;
}
t.clear();
t.push_back(intervals[x][0]);
t.push_back(m);
ans.push_back(t);
i=y-1;
}
return ans;
}
};