【LeetCode】Merge Intervals
題目
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].
分析
先對intervals集合按start從小到大排序,last變量用於保存可能插入到結果集中的元素,遍歷每一個集合中的元素,如果符合合併的條件,將last和當前元素合併,並重新賦值給last,此時last仍然具有合併的潛力;如果不符合合併的條件,則將last放入結果集中,並把當前元素賦值給last,成爲一個新的潛在具有合併性的元素。
自定義比較類
// 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 static final Comparator<Interval> BY_START = new ByStart();
private static class ByStart implements Comparator<Interval> {
@Override
public int compare(Interval o1, Interval o2) {
return o1.start - o2.start;
}
}
主方法
public List<Interval> merge(List<Interval> intervals) {
ArrayList<Interval> result = new ArrayList<Interval>();
if (intervals == null || intervals.size() == 0) {
return result;
}
//按Interval的start對intervals排序
Collections.sort(intervals, BY_START);
Interval last = intervals.get(0);
for (int i = 1; i < intervals.size(); i++) {
Interval temp = intervals.get(i);
if (canMerge(last, temp)) {
if (last.end <= temp.end) {
last = new Interval(last.start, temp.end);
}
//另外一種情況last保持不變
} else {
result.add(last);
last = intervals.get(i);
}
}
result.add(last);
return result;
}
private boolean canMerge(Interval item1, Interval item2) {
if (item1.end >= item2.start) {
return true;
}
return false;
}