在给定的一组间隔里插入一个新的间隔

问题：

给定多个不重合的间隔（interval），比如 [3, 8] [ 12, 17] [22, 29], 然后插入一个新的间隔， 比如 [ 5, 15], 求最后不重合的间隔。比如，起初的不重合间隔变成 [3, 17] [22, 29].

分析：

首先，我们要知道怎么确定两个间隔是重合的。比如[1, 5] [3, 7] 是重合的，[1, 5] [6, 7] 是不重合的。那么，观察后发现，如果两个间隔的起始值的最大值小于两个间隔的结束值的最小值，那么，这两个间隔一定重合。代码如下：

/* * s1: start point of interval 1 * e1: end point of interval 1 * s2: start point of interval 2 * e2: end point of interval 2 */public boolean overlap(int s1, int e1, int s2, int e2) {return Math.max(s1, s2) < Math.min(e1, e2);}

假定给定的间隔按照起始值已经排序了，就是说第一个间隔的起始值是所有间隔起始值中最小的，最后一个间隔的起始值是所有间隔起始值中最大的。当插入一个新的间隔时，我们只需要判断它是否和每一个间隔相交，如果不相交，保持不变，如果相交，则把新的间隔求出来，再继续和后面的间隔进行判断。代码如下：

public class Test {//check whether the two intervals intersectpublic boolean overlap(Interval i1, Interval i2) {return Math.max(i1.start, i2.start) < Math.min(i1.end, i2.end);}//add the new interval to the interval list and return a new interval list without overlappingpublic ArrayList<Interval> add(ArrayList<Interval> list, Interval interval) {ArrayList<Interval> newList = new ArrayList<Interval>();int i = 0;while (i < list.size()) {//no intersectionif (overlap(list.get(i), interval) == false) {newList.add(list.get(i));i++;} else {//get the new intervalwhile (i < list.size() && overlap(list.get(i), interval) == true) {int start = Math.min(list.get(i).start, interval.start);int end = Math.max(list.get(i).end, interval.end);interval = new Interval(start, end);i++;}newList.add(interval);}}return newList;}public static void main(String[] args) {ArrayList<Interval> list = new ArrayList<Interval>();list.add(new Interval(3, 8));list.add(new Interval(22, 29));list.add(new Interval(12, 17));list.add(new Interval(40, 47));Collections.sort(list);Interval interval = new Interval(40, 48);Test test = new Test();ArrayList<Interval> newList = test.add(list, interval);for(Interval inter : newList) {System.out.println("[" + inter.start + ", " + inter.end + "]");}}}class Interval implements Comparable<Interval>{int start;int end;Interval(int start, int end) {this.start = start;this.end = end;}public int compareTo(Interval o) {if (start > ((Interval)o).start || (start == ((Interval)o).start && end > ((Interval)o).end)) {return 1;} else if (start == ((Interval)o).start && end == ((Interval)o).end) {return 0;}return -1;}}

Another solution

/** * 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 ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {        ArrayList<Interval> listAfterMerged = new ArrayList<Interval>();        if (intervals == null || intervals.size() == 0) {            listAfterMerged.add(newInterval);            return listAfterMerged;        }        boolean isAdded = false;                for (int i = 0; i < intervals.size(); i++) {            Interval interv =  intervals.get(i);            if (isAdded == false) {                if (interv.start < newInterval.start) {                    addInterval(listAfterMerged, interv);                } else {                    addInterval(listAfterMerged, newInterval);                    isAdded = true;                    addInterval(listAfterMerged, interv);                }            } else {                addInterval(listAfterMerged, interv);            }        }        if (isAdded == false) {            addInterval(listAfterMerged, newInterval);        }        return listAfterMerged;    }        public void addInterval(ArrayList<Interval> listAfterMerged, Interval newInterval) {        if (listAfterMerged.size() == 0) {            listAfterMerged.add(newInterval);        }         Interval interv =  listAfterMerged.get(listAfterMerged.size() - 1);        if (interv.end >= newInterval.start) {            interv.end = (int) Math.max(interv.end, newInterval.end);        } else {            listAfterMerged.add(newInterval);        }    }            boolean isOverlap(Interval i1, Interval i2) {        if (Math.max(i1.start, i2.start) <= Math.min(i1.end, i2.end)) return true;        return false;    }}

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