[LeetCode] 57. Insert Interval 插入区间

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:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

56. Merge Intervals的拓展，给定一个无重叠的区间集合，以开始时间排序好了，插入一个新的区间，保证集合区间仍然有序且不重叠（如果有必要的话合并区间）。

新建一个数组，先把结尾小于新区间开始的区间写入数组，然后对于后边的和新区间进行合并，生成新的区间，直到新区间的结束小于后边区间的开始，写入这个合并后的新区间和后边剩余的区间到新数组。还有一种做法是，这里要插入一个区间，就要比较新区间和就的区间列表中的区间，如果不需要合并，就直接插入。如果需要合并，则要合并新区间，并删除不需要的区间。

Java：

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public class Solution {     public List<Interval> insert(List<Interval> intervals, Interval newInterval) {         if (newInterval == null || intervals == null) {             return intervals;         }           List<Interval> results = new ArrayList<Interval>();         int insertPos = 0;           for (Interval interval : intervals) {             if (interval.end < newInterval.start) {                 results.add(interval);                 insertPos++;             } else if (interval.start > newInterval.end) {                 results.add(interval);             } else {                 newInterval.start = Math.min(interval.start, newInterval.start);                 newInterval.end = Math.max(interval.end, newInterval.end);             }         }           results.add(insertPos, newInterval);           return results;     } }

Python:

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class Solution(object):     def insert(self, intervals, newInterval):         result = []         i = 0         while i < len(intervals) and newInterval.start > intervals[i].end:             result += intervals[i],             i += 1         while i < len(intervals) and newInterval.end >= intervals[i].start:             newInterval = Interval(min(newInterval.start, intervals[i].start), \                                    max(newInterval.end, intervals[i].end))             i += 1         result += newInterval,         result += intervals[i:]         return result

Python：

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class Solution:     def insert(self, intervals, newInterval):         results = []         insertPos = 0         for interval in intervals:             if interval.end < newInterval.start:                 results.append(interval)                 insertPos += 1             elif interval.start > newInterval.end:                 results.append(interval)             else:                 newInterval.start = min(interval.start, newInterval.start)                 newInterval.end = max(interval.end, newInterval.end)         results.insert(insertPos, newInterval)         return results

C++:

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class Solution { public:     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {         vector<Interval> res = intervals;         int i = 0, overlap = 0, n = res.size();         while (i < n) {             if (newInterval.end < res[i].start) break;              else if (newInterval.start > res[i].end) {}             else {                 newInterval.start = min(newInterval.start, res[i].start);                 newInterval.end = max(newInterval.end, res[i].end);                 ++overlap;             }             ++i;         }         if (overlap > 0) res.erase(res.begin() + i - overlap, res.begin() + i);         res.insert(res.begin() + i - overlap, newInterval);         return res;     } };

C++:

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class Solution { public:     vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {         size_t i = 0;         vector<Interval> result;         // Insert intervals appeared before newInterval.         while (i < intervals.size() && newInterval.start > intervals[i].end) {             result.emplace_back(intervals[i++]);         }           // Merge intervals that overlap with newInterval.         while (i < intervals.size() && newInterval.end >= intervals[i].start) {             newInterval = {min(newInterval.start, intervals[i].start),                 max(newInterval.end, intervals[i].end)};             ++i;         }         result.emplace_back(newInterval);           // Insert intervals appearing after newInterval.         result.insert(result.end(), intervals.cbegin() + i, intervals.cend());         return result;     } };

　　

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[LeetCode] 56. Merge Intervals 合并区间

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