leetcode hard模式专杀之57. Insert Interval

这题思路 不算难，但同样是要注意边界条件， high level的思路是如果新插入的interval没有任何overlap，那就直接插入，否则，就通过如下方式合并：删掉所有的有overlap的interval，再插入一个新的interval

那么问题来了，如何确定哪些是overlap的interval, 新的interval又是从哪儿到哪儿？

要回答这两个问题，通过两次循环遍历足矣，上代码：

/**
 * 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 int compare(Interval interval, int pos){
        if(pos<interval.start){
            // point to the left
            return -1;
        }else if(pos<=interval.end){
            // in the interval
            return 0;
        }else{
            // point to the right
            return 1;
        }
    }
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        int startPos = 0;
        int endPos = 0;

        int mergeStart = newInterval.start;
        int mergeEnd = newInterval.end;

        boolean rightOverlap = false;

        int i = 0;
        for(i = 0 ; i<intervals.size(); i++){
            Interval current = intervals.get(i);
            int comp = compare(current, newInterval.start);
            if(comp==1){
                continue;
            }else if(comp==0){
                startPos = i;
                mergeStart = current.start;
                break;
            }else{
                startPos = i;
                mergeStart = newInterval.start;
                break;
            }
        }

        // move all the way to the end
        if(i==intervals.size()){
            intervals.add(intervals.size(), newInterval);
            return intervals;
        }

        for(i = 0 ; i<intervals.size(); i++){
            Interval current = intervals.get(i);
            int comp = compare(current, newInterval.end);
            if(comp==1){
                continue;
            }else if(comp==0){
                endPos = i;
                mergeEnd = current.end;
                rightOverlap = true;
                break;
            }else{
                endPos = i;
                mergeEnd = newInterval.end;
                break;
            }
        }

        if(i==intervals.size()){
            endPos = i;
        }

        // has overlap
        if(rightOverlap){
            //delete from startPos to endPos and insert a new interval
            for(i =0 ; i< endPos-startPos+1; i++){
                intervals.remove(startPos);
            }
            intervals.add(startPos, new Interval(mergeStart, mergeEnd));
        }else{
            // delete from startPos to endPos-1 and insert a new interval
            for(i=0; i<endPos-startPos; i++){
                intervals.remove(startPos);
            }
            intervals.add(startPos, new Interval(mergeStart, mergeEnd));
        }
        return intervals;
    }
}