leetcode56- Merge Intervals- medium

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].

先写comparator按照interval的开始时间出现早晚排好序。接着开始遍历每个interval，如果和上一个有交叉(crt.start <= prev.end)，那就改上一个的end为更持久的一个，如果和上一个没有交叉，那就把之前记录的间隔推入结果中，重新开始新的检查。

细节：发现交叉的时候不是直接把尾赋值为当前的，而是要赋值更长的一个。这样才能避免包含关系下出现逻辑错误。

1.O(n)空间：

/**
 * 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; }
 * }
 */
class Solution {
    
    private Comparator<Interval> comparator = new Comparator<Interval>(){
        public int compare(Interval i1, Interval i2) {
            return i1.start - i2.start;
        }
    };
    
    public List<Interval> merge(List<Interval> intervals) {
        
        List<Interval> result = new ArrayList<Interval>();
        if (intervals == null || intervals.size() == 0) {
            return result;
        }
        
        Collections.sort(intervals, comparator);
        
        int start = intervals.get(0).start;
        int end = intervals.get(0).end;
        for (int i = 1; i < intervals.size(); i++) {
            Interval crt = intervals.get(i);
            if (crt.start <= end) {
                end = Math.max(end, crt.end);
            } else {
                result.add(new Interval(start, end));
                start = crt.start;
                end = crt.end;
            }
        }
        result.add(new Interval(start, end));
        
        return result;
    }    
}

2.O(1)空间，原地合并

/**
 * 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; }
 * }
 */
class Solution {
    
    private Comparator<Interval> comparator = new Comparator<Interval>(){
        public int compare(Interval i1, Interval i2) {
            return i1.start - i2.start;
        }
    };
    
    public List<Interval> merge(List<Interval> intervals) {
        
        List<Interval> result = new ArrayList<Interval>();
        if (intervals == null || intervals.size() == 0) {
            return result;
        }
        
        Collections.sort(intervals, comparator);
        
        Interval last = intervals.get(0);
        for (int i = 1; i < intervals.size(); i++) {
            Interval crt = intervals.get(i);
            if (crt.start <= last.end) {
                last.end = Math.max(last.end, crt.end);
            } else {
                result.add(last);
                last = crt;
            }
        }
        result.add(last);
        
        return result;
    }    
}