题目:

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings.

For example,
Given [[0, 30],[5, 10],[15, 20]],
return false.

链接: http://leetcode.com/problems/meeting-rooms/

题解:

一开始以为是跟Course Schedule一样,仔细读完题目以后发现只要sort一下就可以了。写得还不够精简,需要好好研究一下Java8的lambda表达式。

Time Complexity - O(nlogn), Space Complexity - O(1)

/**
* 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 boolean canAttendMeetings(Interval[] intervals) {
if(intervals == null || intervals.length == 0)
return true;
Arrays.sort(intervals, new Comparator<Interval>(){
public int compare(Interval t1, Interval t2) {
if(t1.start != t2.start)
return t1.start - t2.start;
else
return t1.end - t2.end;
}
}); for(int i = 1; i < intervals.length; i++) {
if(intervals[i].start < intervals[i - 1].end)
return false;
} return true;
}
}

二刷:

也是先对interval数组进行先startdata再enddate的排序,之后一次遍历数组来看是否intervals[i].start < intervals[i - 1].end。

用lambda表达式以后发现好慢

Java:

Time Complexity - O(nlogn), Space Complexity - O(1)

/**
* 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 boolean canAttendMeetings(Interval[] intervals) {
if (intervals == null) {
return true;
}
Arrays.sort(intervals, (Interval i1, Interval i2) -> i1.start != i2.start ? i1.start - i2.start : i1.end - i2.end);
for (int i = 1; i < intervals.length; i++) {
if (intervals[i].start < intervals[i - 1].end) {
return false;
}
}
return true;
}
}

三刷:

Java:

/**
* 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 boolean canAttendMeetings(Interval[] intervals) {
if (intervals == null || intervals.length == 0) return true;
Arrays.sort(intervals, (Interval i1, Interval i2) -> i1.start != i2.start ? i1.start - i2.start : i1.end - i2.end);
for (int i = 1; i < intervals.length; i++) {
if (intervals[i].start < intervals[i - 1].end) return false;
}
return true;
}
}

Reference:

https://leetcode.com/discuss/50912/ac-clean-java-solution

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