[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：

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:

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：

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++:

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++:

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;
    }
};

　　

类似题目：

[LeetCode] 56. Merge Intervals 合并区间

All LeetCode Questions List 题目汇总