合并所有覆盖的区间,一道对逻辑思维要求比较高的题。
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].
http://oj.leetcode.com/problems/insert-interval/
解题报告:
新区间是新输入的线段,对原有的线段ArrayList进行遍历。
考虑以下情况:
如果新区间的end < 当前区间的start,不用找下去了,把新区间插入到当前区间的前面,然后返回。
如果当前区间的end小于新区间的start,继续遍历找下一个区间。
如果当前区间和新区间发生重合,则start取两者最小的start,end取两者最大的end,生成一个新的区间。
继续遍历。
如果遍历一直到末尾也没发生区间重合,就把新区间插入到原来ArrayList的末尾。
里面用到了几个通过iterator操作ArrayList的函数,有空慢慢讲解一下。
AC代码:
public class Solution {
public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
if(intervals==null||newInterval==null) {
return intervals;
}
if(intervals.size()==0) {
intervals.add(newInterval);
return intervals;
}
ListIterator<Interval> it = intervals.listIterator();
while(it.hasNext()) {
Interval tmpInterval = it.next();
if(newInterval.end<tmpInterval.start) {
it.previous();
it.add(newInterval);
return intervals;
} else {
if(tmpInterval.end<newInterval.start) {
continue;
} else {
newInterval.start = Math.min(tmpInterval.start, newInterval.start);
newInterval.end = Math.max(tmpInterval.end, newInterval.end);
it.remove();
}
}
}
intervals.add(newInterval);
return intervals;
}
}
LeetCode Insert Interval,布布扣,bubuko.com
LeetCode Insert Interval
原文:http://blog.csdn.net/worldwindjp/article/details/21612731