56. Merge Intervals

题目：

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].

链接：https://leetcode.com/problems/merge-intervals/#/description

4/18/2017

22%， 34ms

要好好研究Java的sort各种知识，比如comparator, comparable等等，这道题的sort部分很多是参考课件

注意的问题：

1. Arrays.sort()是个array用的，List直接用List.sort()

2. 如何实现compare函数，comparator接口，以及comparator的类

public class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        intervals.sort(new IntervalComparator());
        List<Interval> ret = new ArrayList<Interval>();
        int i = -1;
        for (Interval interval: intervals) {
            if (i == -1) {
                ret.add(interval);
                i++;
            } else {
                if (ret.get(i).end < interval.start) {
                    ret.add(interval);
                    i++;
                } else if (ret.get(i).end < interval.end) {
                    Interval tmp = new Interval();
                    tmp.start = ret.get(i).start;
                    tmp.end = interval.end;
                    ret.set(i, tmp);
                }
            }
        }
        return ret;
    }
    private static class IntervalComparator implements Comparator<Interval> {
        public int compare(Interval a, Interval b) {
            if      (a.start < b.start) return -1;
            else if (a.start > b.start) return +1;
            else if (a.end < b.end) return -1;
            else if (a.end > b.end) return +1;
            else                    return  0;
        }
    }
}

别人的算法：

这个想法跟我一样，更新ret部分我觉得反而不够清晰，不过可以看如何实现比较

https://discuss.leetcode.com/topic/4319/a-simple-java-solution

public List<Interval> merge(List<Interval> intervals) {
    if (intervals.size() <= 1)
        return intervals;
    
    // Sort by ascending starting point using an anonymous Comparator
    intervals.sort((i1, i2) -> Integer.compare(i1.start, i2.start));
    
    List<Interval> result = new LinkedList<Interval>();
    int start = intervals.get(0).start;
    int end = intervals.get(0).end;
    
    for (Interval interval : intervals) {
        if (interval.start <= end) // Overlapping intervals, move the end if needed
            end = Math.max(end, interval.end);
        else {                     // Disjoint intervals, add the previous one and reset bounds
            result.add(new Interval(start, end));
            start = interval.start;
            end = interval.end;
        }
    }
    
    // Add the last interval
    result.add(new Interval(start, end));
    return result;
}

可以用Collections.sort

Collections.sort(intervals, new Comparator<Interval>() {
        public int compare(Interval i1, Interval i2) {
            if (i1.start != i2.start) {
                return i1.start - i2.start;
            }
            return i1.end - i2.end;
        }
    });

另外好像比较时候不需要很精细的比较end，这道题里比较start以及足够

https://discuss.leetcode.com/topic/20263/c-10-line-solution-easing-understanding

vector<Interval> merge(vector<Interval>& ins) {
    if (ins.empty()) return vector<Interval>{};
    vector<Interval> res;
    sort(ins.begin(), ins.end(), [](Interval a, Interval b){return a.start < b.start;});
    res.push_back(ins[0]);
    for (int i = 1; i < ins.size(); i++) {
        if (res.back().end < ins[i].start) res.push_back(ins[i]);
        else
            res.back().end = max(res.back().end, ins[i].end);
    }
    return res;
}

Python

https://discuss.leetcode.com/topic/17178/7-lines-easy-python

def merge(self, intervals):
    out = []
    for i in sorted(intervals, key=lambda i: i.start):
        if out and i.start <= out[-1].end:
            out[-1].end = max(out[-1].end, i.end)
        else:
            out += i,
    return out

更多讨论：

https://discuss.leetcode.com/category/64/merge-intervals