Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
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].
Example 3:
Input: intervals = [], newInterval = [5,7] Output: [[5,7]]
Example 4:
Input: intervals = [[1,5]], newInterval = [2,3] Output: [[1,5]]
Example 5:
Input: intervals = [[1,5]], newInterval = [2,7] Output: [[1,7]]
Constraints:
0 <= intervals.length <= 104intervals[i].length == 20 <= intervals[i][0] <= intervals[i][1] <= 105intervals is sorted by intervals[i][0] in ascending order.newInterval.length == 20 <= newInterval[0] <= newInterval[1] <= 105struct Solution;
impl Solution {
fn insert(intervals: Vec<Vec<i32>>, mut new_interval: Vec<i32>) -> Vec<Vec<i32>> {
let mut res = vec![];
for interval in intervals {
if interval[0] < new_interval[0] {
if interval[1] < new_interval[0] {
res.push(interval);
} else {
new_interval[0] = new_interval[0].min(interval[0]);
new_interval[1] = new_interval[1].max(interval[1]);
}
} else {
if interval[0] > new_interval[1] {
res.push(interval);
} else {
new_interval[0] = new_interval[0].min(interval[0]);
new_interval[1] = new_interval[1].max(interval[1]);
}
}
}
if let Err(i) = res.binary_search_by_key(&new_interval[0], |v| v[0]) {
res.insert(i, new_interval);
}
res
}
}
#[test]
fn test() {
let intervals = vec_vec_i32![[1, 3], [6, 9]];
let new_interval = vec![2, 5];
let res = vec_vec_i32![[1, 5], [6, 9]];
assert_eq!(Solution::insert(intervals, new_interval), res);
}