57 Insert Interval 插入区间
Description:
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:
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].
NOTE:
input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
题目描述:
给出一个无重叠的 ，按照区间起始端点排序的区间列表。
在列表中插入一个新的区间，你需要确保列表中的区间仍然有序且不重叠（如果有必要的话，可以合并区间）。
示例 :
示例 1:
输入: intervals = [[1,3],[6,9]], newInterval = [2,5]
输出: [[1,5],[6,9]]
示例 2:
输入: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
输出: [[1,2],[3,10],[12,16]]
解释: 这是因为新的区间 [4,8] 与 [3,5],[6,7],[8,10] 重叠。
思路:
- 遍历 intervals数组, 将 newInterval左边的区间加入结果数组
- 合并区间, 将合并的结果加入结果数组
- 将 intervals数组剩下的元素加入结果数组
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
vector> insert(vector>& intervals, vector& newInterval)
{
int i = 0;
vector> result;
while (i < intervals.size() and newInterval[0] > intervals[i][1]) result.push_back(intervals[i++]);
while (i < intervals.size() and newInterval[1] >= intervals[i][0])
{
newInterval[0] = min(newInterval[0], intervals[i][0]);
newInterval[1] = max(newInterval[1], intervals[i++][1]);
}
result.push_back(newInterval);
while (i < intervals.size()) result.push_back(intervals[i++]);
return result;
}
};
Java:
class Solution {
public int[][] insert(int[][] intervals, int[] newInterval) {
int i = 0, j = 0, result[][] = new int[intervals.length == 0 ? 1 : intervals.length + 1][2];
while (i < intervals.length && newInterval[0] > intervals[i][1]) {
result[j][0] = intervals[i][0];
result[j++][1] = intervals[i++][1];
}
while (i < intervals.length && newInterval[1] >= intervals[i][0]) {
newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
newInterval[1] = Math.max(newInterval[1], intervals[i++][1]);
}
result[j][0] = newInterval[0];
result[j++][1] = newInterval[1];
while (i < intervals.length) {
result[j][0] = intervals[i][0];
result[j++][1] = intervals[i++][1];
}
return Arrays.copyOf(result, j);
}
}
Python:
class Solution:
def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
result, i = [], 0
while i < len(intervals) and newInterval[0] > intervals[i][1]:
result.append(intervals[i])
i += 1
while i < len(intervals) and newInterval[1] >= intervals[i][0]:
newInterval[0], newInterval[1] = min(newInterval[0], intervals[i][0]), max(newInterval[1], intervals[i][1])
i += 1
result.append(newInterval)
while i < len(intervals):
result.append(intervals[i])
i += 1
return result