56. Merge Intervals

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本文详细介绍了一种用于合并重叠区间的算法实现，通过排序和遍历的方式，有效地将输入的一系列区间合并为不重叠的最小集合。示例包括了具体的输入输出对比，帮助理解算法的工作原理。

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56. Merge Intervals

Medium

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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.

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class Solution {

public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        sort(intervals.begin(),intervals.end());//按照左区间边界从小到大排序
        int n=intervals.size();
        if(n<=1) return intervals;
        int left=intervals[0][0];
        int right=intervals[0][1];
        vector<vector<int>> rt;
        for(int i=1;i<n;i++){
        	int templeft=intervals[i][0];//依次比较边界关系并push
        	int tempright=intervals[i][1];
        	if(templeft<=right){
                if(tempright>=right){
                    left=left;
        		    right=tempright;
                }else{
                    left=left;
                    right=right;
                }
        	}else{
        		vector<int> tmp;
        		tmp.push_back(left);
        		tmp.push_back(right);
        		rt.push_back(tmp);
        		left=templeft;
        		right=tempright;
                
        	}
            if(i==n-1){
                vector<int> tp;
                tp.push_back(left);
                tp.push_back(right);
                rt.push_back(tp);
            }
        }
        return rt;
    }
public://这个是多余的
    bool cmp(vector<int> a,vector<int> b){
        return a[0]<b[0];
    }
};