# 57. Insert Interval

You are given an array of non-overlapping intervals `intervals` where `intervals[i] = [starti, endi]` represent the start and the end of the `ith` interval and `intervals` is sorted in ascending order by `starti`. You are also given an interval `newInterval = [start, end]` that represents the start and end of another interval.

Insert `newInterval` into `intervals` such that `intervals` is still sorted in ascending order by `starti` and `intervals` still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return `intervals` after the insertion.

Example 1:

``````Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
``````

Example 2:

``````Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
``````

Constraints:

• `0 <= intervals.length <= 10^4`
• `intervals[i].length == 2`
• `0 <= starti <= endi <= 10^5`
• `intervals` is sorted by `starti` in ascending order.
• `newInterval.length == 2`
• `0 <= start <= end <= 10^5`

``````class Solution {
public:
vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
vector<vector<int>> res;
int n = intervals.size(), cur = 0;
while (cur < n && intervals[cur][1] < newInterval[0]) {
res.push_back(intervals[cur++]);
}
while (cur < n && intervals[cur][0] <= newInterval[1]) {
newInterval[0] = min(newInterval[0], intervals[cur][0]);
newInterval[1] = max(newInterval[1], intervals[cur][1]);
++cur;
}
res.push_back(newInterval);
while (cur < n) {
res.push_back(intervals[cur++]);
}
return res;
}
};
``````

``````class Solution {
public:
vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
vector<vector<int>> res;
int n = intervals.size(), cur = 0;
for (int i = 0; i < n; ++i) {
if (intervals[i][1] < newInterval[0]) {
res.push_back(intervals[i]);
++cur;
} else if (intervals[i][0] > newInterval[1]) {
res.push_back(intervals[i]);
} else {
newInterval[0] = min(newInterval[0], intervals[i][0]);
newInterval[1] = max(newInterval[1], intervals[i][1]);
}
}
res.insert(res.begin() + cur, newInterval);
return res;
}
};
``````

``````class Solution {
public:
vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
vector<vector<int>> res;
int n = intervals.size(), cur = 0, i = 0;
while (i < n) {
if (intervals[i][1] < newInterval[0]) {
res.push_back(intervals[i]);
++cur;
} else if (intervals[i][0] > newInterval[1]) {
res.push_back(intervals[i]);
} else {
newInterval[0] = min(newInterval[0], intervals[i][0]);
newInterval[1] = max(newInterval[1], intervals[i][1]);
}
++i;
}
res.insert(res.begin() + cur, newInterval);
return res;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/57

Range Module

Merge Intervals

Count Integers in Intervals

https://leetcode.com/problems/insert-interval/

https://leetcode.com/problems/insert-interval/discuss/21669/Easy-and-clean-O(n)-C++-solution

https://leetcode.com/problems/insert-interval/discuss/21602/Short-and-straight-forward-Java-solution

https://leetcode.com/problems/insert-interval/discuss/21676/Clean-and-short-Java-solution-with-explanation

LeetCode All in One 题目讲解汇总(持续更新中…)

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