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  • Insert Intervals

     

    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].

    /**
     * Definition for an interval.
     * public class Interval {
     *     int start;
     *     int end;
     *     Interval() { start = 0; end = 0; }
     *     Interval(int s, int e) { start = s; end = e; }
     * }
     */
    public class Solution {
        public ArrayList<Interval> insert(ArrayList<Interval> intervals,
                Interval newInterval) {
                    
            ArrayList<Interval> result = new ArrayList<Interval>();
            
            for(int i = 0; i < intervals.size(); i ++){
                Interval temp = intervals.get(i);
                
                if(temp.start> newInterval.end){
                    result.add(newInterval);
                    for(int j = i; j < intervals.size(); j++){
                        result.add(intervals.get(j));
                    }
                    return result;
                } else if( newInterval.start > temp.end){
                    result.add(temp);
                    continue;
                } else{
                    // Attention : method Math.min and Math.max
                    newInterval.start = Math.min(newInterval.start, temp.start);
                    newInterval.end = Math.max(newInterval.end, temp.end);
                }
            }
            /*easy to forget. The situation that newInterval.start is the biggest */
            result.add(newInterval);
            return result;
        }
    }
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  • 原文地址:https://www.cnblogs.com/RazerLu/p/3532267.html
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