Insert Interval

描述

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].

分析

代码

struct Interval {
    int start;
    int end;
    Interval() : start(0), end(0) { }
    Interval(int s, int e) : start(s), end(e) { }
};

// Insert Interval
// 时间复杂度O(n),空间复杂度O(1)
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        for (int i = 0; i < intervals.size();) {
            final Interval cur = intervals.get(i);
            if (newInterval.end < cur.start) {
                intervals.add(i, newInterval);
                return intervals;
            } else if (newInterval.start > cur.end) {
                ++i;
                continue;
            } else {
                newInterval.start = Math.min(newInterval.start, cur.start);
                newInterval.end = Math.max(newInterval.end, cur.end);
                intervals.remove(i);
            }
        }
        intervals.add(newInterval);
        return intervals;
    }
}

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