leetcode题解-57. Insert Interval

题目：

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].

本题是对于给定的non-overlapping intervals和一个新的intervals，要将新的插进去并进行合并，然后我们可以想到无非就是分别找到newInterval.start和newInterval.end插入的位置即可。因此我们可以使用二插搜索的方法分别找到其茶如位置，然后将器进行合并，代码入下所示：

public List<Interval> insert1(List<Interval> intervals, Interval newInterval) {
int n = intervals.size();
if (n == 0) {
intervals.add(newInterval);
return intervals;
}

int low = 0, high = n - 1, mid = 0;
int temp, target = newInterval.start;
while (low <= high) {
mid = (low + high) / 2;
temp = intervals.get(mid).start;
if (temp == target)
break;
if (temp < target)
low = mid + 1;
else
high = mid - 1;
}

// insIdx = the index where new interval to be inserted
int insIdx = (low <= high) ? mid : low;
Interval pre = (insIdx == 0) ? null : intervals.get(insIdx - 1);
// 0<=insIdx<=n, pre=[insIdx-1], pre.start<new.start

low = insIdx;
high = n - 1;
target = newInterval.end;
while (low <= high) {
mid = (low + high) / 2;
temp = intervals.get(mid).end;
if (temp == target)
break;
if (temp < target)
low = mid + 1;
else
high = mid - 1;
}

// nxtIdx= the next index after the inserted new interval
int nxtIdx = (low <= high) ? mid : low;
Interval nxt = (nxtIdx == n) ? null : intervals.get(nxtIdx);
// insIdx<=nxtIdx<=n, nxt=[nxtIdx], nxt.end>=new.end

// [0]...[insIdx-1] <--> [insIdx]...[nxtIdx-1][nxtIdx]...

intervals.subList(insIdx, nxtIdx).clear();

// check whether newInterval can be merged with pre or nxt
boolean isMerged = false, isMerged2 = false;
if (insIdx > 0 && pre.end >= newInterval.start) {
pre.end = Math.max(pre.end, newInterval.end);
isMerged = true;
}

if (nxtIdx < n && newInterval.end >= nxt.start) {
nxt.start = Math.min(nxt.start, newInterval.start);
isMerged2 = isMerged;
isMerged = true;
}

if (!isMerged) {
intervals.add(insIdx, newInterval);
return intervals;
}

// merged with pre or nxt or both, deal with the both case
if (isMerged2 && pre.end >= nxt.start) {
nxt.start = pre.start; // pre.start < new.start, nxt.start;
intervals.remove(insIdx - 1); // remove pre
}

return intervals;
}

另外一种思路就是我们遍历链表，将其分为三部分，第一部分是比newInterval.start小的，我们直接将其插入到结果之中，第二部分是需要合并的，我们生成一个新的Interval并插入，第三部分是后面的比end打的，我们也直接插入结果即可。代码入小所示：

//53.6%
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> result = new ArrayList<>();
int i = 0;
// add all the intervals ending before newInterval starts
while (i < intervals.size() && intervals.get(i).end < newInterval.start)
result.add(intervals.get(i++));
// merge all overlapping intervals to one considering newInterval
while (i < intervals.size() && intervals.get(i).start <= newInterval.end) {
newInterval = new Interval( // we could mutate newInterval here also
Math.min(newInterval.start, intervals.get(i).start),
Math.max(newInterval.end, intervals.get(i).end));
i++;
}
result.add(newInterval); // add the union of intervals we got
// add all the rest
while (i < intervals.size()) result.add(intervals.get(i++));
return result;
}