Problem 56: Merge Intervals

https://leetcode.com/problems/merge-intervals/

思路

• 首先对所有的 interval 进行排序，排序的原则按照 start 进行

• 然后遍历整个数组。如果中间有 overlap 的话，合并 end，然后删掉 next interval。

复杂度

• sort 花费 O(nlogn) 的时间，便利（merge）花费O(n)的时间

• Time: **O(nlogn)**

/**
 * 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 List<Interval> merge(List<Interval> intervals) {
        Collections.sort(intervals, new Comparator<Interval>() {
            public int compare(Interval i1, Interval i2) {
                return i1.start - i2.start;
            }
        });

        int index = 0;
        while (index < intervals.size() - 1) {
            Interval curr = intervals.get(index);
            Interval next = intervals.get(index + 1);
            if (next.start <= curr.end) {
                curr.end = Math.max(curr.end, next.end);
                intervals.remove(index + 1);
            } else { 
                index++;
            }
        }

        return intervals;
    }
}

易错点

1. Comparator 的应用以及 Collections 的排序

2. 遍历数组的时候别忘了边界

   while (index < intervals.size() - 1)

3. Comparator 的 Lambda 表达式

   Collections.sort(intervals, (i1, i2) -> (i1.start - i2.start));