simulation/merge-intervals

Merge Intervals

描述

Given a collection of intervals, merge all overlapping intervals.

For example, Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18]

分析

复用一下 Insert Intervals 的解法即可，创建一个新的 interval 集合，然后每次从旧的里面取一个 interval 出来，然后插入到新的集合中。

代码

// Merge Interval
// 复用一下Insert Intervals的解法即可
// 时间复杂度O(n1+n2+...)，空间复杂度O(1)
public class Solution {
    public int[][] merge(int[][] intervals) {
        ArrayList<int[]> list = new ArrayList<>();
        for (int[] newInterval : intervals) {
            insert(list, newInterval);
        }
        return list.toArray(new int[0][2]);
    }

    private void insert(ArrayList<int[]> intervals, int[] newInterval) {
        for (int i = 0; i < intervals.size();) {
            final int[] cur = intervals.get(i);
            if (newInterval[1] < cur[0]) {
                intervals.add(i, newInterval);
                return;
            } else if (newInterval[0] > cur[1]) {
                ++i;
            } else {
                newInterval[0] = Math.min(newInterval[0], cur[0]);
                newInterval[1] = Math.max(newInterval[1], cur[1]);
                intervals.remove(i);
            }
        }
        intervals.add(newInterval);
    }
}

// Merge Interval
//复用一下Insert Intervals的解法即可
// 时间复杂度O(n1+n2+...)，空间复杂度O(1)
class Solution {
public:
    vector<Interval> merge(vector<Interval> &intervals) {
        vector<Interval> result;
        for (int i = 0; i < intervals.size(); i++) {
            insert(result, intervals[i]);
        }
        return result;
    }
private:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval>::iterator it = intervals.begin();
        while (it != intervals.end()) {
            if (newInterval.end < it->start) {
                intervals.insert(it, newInterval);
                return intervals;
            } else if (newInterval.start > it->end) {
                it++;
                continue;
            } else {
                newInterval.start = min(newInterval.start, it->start);
                newInterval.end = max(newInterval.end, it->end);
                it = intervals.erase(it);
            }
        }
        intervals.insert(intervals.end(), newInterval);
        return intervals;
    }
};

相关题目

• Insert Interval
• Employee Free Time