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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].
这道题和上面的一道题是类似题，就是给定数个排序好的数组和一个单独的数组，将单独的数组插入那个数组集合中，并且合并相关的集合。
第一次写的就是找到开始插入的地方begin和插入结束的地方last，然后删除并且合并，但是结果并不是很理想。
/** * 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 List<Interval> insert(List<Interval> intervals, Interval newInterval) { int start = newInterval.start,end = newInterval.end,len = intervals.size(); int begin = 0, last = len-1; if( len == 0){ intervals.add(newInterval); return intervals; } while( begin<len && intervals.get(begin).end < start ) begin++; if( begin == len){ intervals.add(newInterval); return intervals; } while( last>=0 && intervals.get(last).start > end ) last--; //from begin to last if( begin > last){ intervals.add(begin, newInterval); return intervals; } Interval ans = new Interval(); ans.start = intervals.get(begin).start>newInterval.start?newInterval.start:intervals.get(begin).start; ans.end = intervals.get(last).end>newInterval.end?intervals.get(last).end:newInterval.end; intervals.set(begin, ans); for( int i = begin+1;i<=last;i++){ intervals.remove(begin+1); } return intervals; } }
然后做了一点改变，就是last不从最后开始算起，而是从begin开始的地方算起，结果速度更慢了＝ ＝。。。。。
/** * 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 List<Interval> insert(List<Interval> intervals, Interval newInterval) { int start = newInterval.start, end = newInterval.end, len = intervals.size(); int begin = 0, last = 0 ; if (len == 0) { intervals.add(newInterval); return intervals; } if ( end < intervals.get(0).start ){ intervals.add(0, newInterval); return intervals; } if ( start > intervals.get(len-1).end){ intervals.add(newInterval); return intervals; } while ( start > intervals.get(begin).end) begin++; last = begin; while ( last < len && end >= intervals.get(last).start ) last++; if( begin == last ){ intervals.add(begin,newInterval); return intervals; } start = intervals.get(begin).start>start?start:intervals.get(begin).start; end = intervals.get(last-1).end>end?intervals.get(last-1).end:end; for ( int i = 1;i<last-begin;i++) intervals.remove(begin); intervals.get(begin).start = start; intervals.get(begin).end = end; return intervals; } }
估计效率如此低的缘故，应该是remove操作造成的，因为每一次remove操作是将该元素删除之后，再将之后的所有元素向前移动一次，所以效率很低。
所以改变直接remove的方式，改为先将后面的集合移动过来，然后再从后向前删除集合。这样效率就很高了。
/** * 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 List<Interval> insert(List<Interval> intervals, Interval newInterval) { int start = newInterval.start, end = newInterval.end, len = intervals.size(); int begin = 0, last = 0 ; if (len == 0) { intervals.add(newInterval); return intervals; } if ( end < intervals.get(0).start ){ intervals.add(0, newInterval); return intervals; } if ( start > intervals.get(len-1).end){ intervals.add(newInterval); return intervals; } while ( start > intervals.get(begin).end) begin++; last = begin; while ( last < len && end >= intervals.get(last).start ) last++; if( begin == last ){ intervals.add(begin,newInterval); return intervals; } intervals.get(begin).start = intervals.get(begin).start>start?start:intervals.get(begin).start; intervals.get(begin).end = intervals.get(last-1).end>end?intervals.get(last-1).end:end; for( int i = 0;i<len-last;i++) intervals.set(begin+1+i,intervals.get(last+i)); for ( int i = 1;i<last-begin;i++) intervals.remove(len-i); return intervals; } }
leetcode 57 Insert Interval ----- java
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原文地址：http://www.cnblogs.com/xiaoba1203/p/5808581.html
