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docs/Leetcode_Solutions/C++/0056._Merge_Intervals.md

+# 56. Merge Intervals
+
+**<font color=red>难度:Medium<font>**
+
+## 刷题内容
+> 原题连接
+
+* https://leetcode.com/problems/merge-intervals/
+
+> 内容描述
+
+```
+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.
+```
+
+> 思路1
+******- 时间复杂度: O(nlgn)******- 空间复杂度: O(1)******
+
+这里我们要先对数组进行排序，按照start进行升序排序。接下来就遍历排序好的数组，如过不相交就将两个区间都存入ans。如果相交就将合并后的区间的并入ans。
+
+```cpp
+/**
+ * Definition for an interval.
+ * struct Interval {
+ *     int start;
+ *     int end;
+ *     Interval() : start(0), end(0) {}
+ *     Interval(int s, int e) : start(s), end(e) {}
+ * };
+ */
+class Solution {
+public:
+    vector<Interval> merge(vector<Interval>& intervals) {
+         vector<Interval> ans;
+        if(!intervals.size())
+            return ans;
+        sort(intervals.begin(),intervals.end(),[](Interval a, Interval b){return a.start < b.start;});
+        for(int i = 0;i < intervals.size() - 1;++i)
+            if(intervals[i + 1].start <= intervals[i].end)
+            {
+                intervals[i + 1].start = intervals[i].start;
+                intervals[i].end = max(intervals[i].end,intervals[i + 1].end);
+                intervals[i + 1].end = intervals[i].end;
+            }
+            else
+                ans.push_back(intervals[i]);
+        ans.push_back(intervals[intervals.size() - 1]);
+        return ans;
+    }
+};
+```
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