57 Insert Interval

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 List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> res = new ArrayList();
if(newInterval == null)
return intervals;
if(newInterval != null && (intervals == null || intervals.size() == 0)) {
res.add(newInterval);
return res;
}
int start = -1;
boolean flag = false;
for(Interval i : intervals) {
if(flag) { // 给定的区间已经被归并，不再与剩余区间做比较
res.add(i);
continue;
}
if(i.start <= newInterval.start && i.end >= newInterval.end) { // 区间内
res.add(i);
flag = true;
continue;
} else if(i.end < newInterval.start || i.start > newInterval.end) { // 区间外
if(i.start > newInterval.end) {
if(start != -1)
res.add(new Interval(start, newInterval.end));
else res.add(newInterval);
flag = true;
}
res.add(i);
continue;
} else { // 跨区间
if(start == -1) // 遇到第一个被跨区间的区间
start = Math.min(i.start, newInterval.start);
if(i.end >= newInterval.end) {
res.add(new Interval(start, i.end));
flag = true;
}
}
}
if(!flag) {
if(start != -1)
res.add(new Interval(start, newInterval.end));
else res.add(newInterval);
}
return res;
}
}

151 / 151 test cases passed.
Status: Accepted
Runtime: 4 ms