本文共 2026 字,大约阅读时间需要 6 分钟。
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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]Output: [[1,5],[6,9]]
Example 2:
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]Output: [[1,2],[3,10],[12,16]]Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10]
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
hard 级别,题目让我们在一系列非重叠的区间中插入一个新的区间。上个区间的题目: 是合并。这个还要复杂些,因为单纯的没有重合的区域,遍历原来的区间位置,在对应位置直接插入就行,重合的不行,重合的区域遇到多个重合的情况,可能要更新为一个新的区间范围,包含原来的区间,再把新的区间加入到结果集。
具体实现思路就是循环并合并,for循环现有区间,判断与新插入的Interval 是否重合
。不重合直接加入到结果集。重合的取新的区间范围,min,max 分别取最小与最大值。在接着判断下一个元素是否可以合并。
public static void main(String[] args) { int[][] intervals ={ {1,2},{3,5},{6,7},{8,10},{12,16} }; int[] newInterval = {4,8}; int[][] res =insert(intervals,newInterval); System.out.println( JSON.toJSON(res)); } public static int[][] insert(int[][] intervals, int[] newInterval) { Listres = new ArrayList (); for(int i=0;i newInterval[1]){ res.add(intervals[i] ); } else{//重叠,进行合并更新interval newInterval[0] = Math.min(newInterval[0] ,intervals[i][0]); newInterval[1] = Math.max(newInterval[1], intervals[i][1]); } }//加入最后一个区间 res.add(newInterval); int[][] temp = res.toArray(new int[0][0]); Arrays.sort(temp, new Comparator (){ @Override public int compare(int[] o1, int[] o2) { // TODO Auto-generated method stub return Integer.compare(o1[0],o2[0]); } }); return temp; }
Runtime: 2 ms, faster than 39.71% of Java online submissions for Insert Interval.
Memory Usage: 41.6 MB, less than 71.88% of Java online submissions for Insert Interval.
最后加了排序,输出可能是乱序的。
因为加了排序,所以时间复杂度O(NlogN). 有时间再看看网上大神是怎么做的。
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