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问题描述
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].
解决思路
分三部分：
1. 把之前的较小区间加入list；
2. 更新重叠的区间的start和end，并加入新的区间；
3. 将剩余的区间加入。
程序
public class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
if (newInterval == null) {
return intervals;
}
List<Interval> merge = new ArrayList<Interval>();
if (intervals.size() == 0) {
merge.add(newInterval);
return merge;
}
int i = 0, n = intervals.size();
// add smaller
while (i < n && intervals.get(i).end < newInterval.start) {
merge.add(intervals.get(i++));
}
// merge
while (i < n && intervals.get(i).start <= newInterval.end) {
newInterval.start = Math.min(newInterval.start, intervals.get(i).start);
newInterval.end = Math.max(newInterval.end, intervals.get(i).end);
++i;
}
merge.add(newInterval);
// add last
while (i < n) {
merge.add(intervals.get(i++));
}
return merge;
}
}
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原文地址：http://www.cnblogs.com/harrygogo/p/4691613.html
