LeetCode Weekly Contest 23 之 539. Minimum Time Difference
2017-03-12 16:23
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LeetCode Weekly Contest 23
赛题
本次周赛主要分为以下4道题:541 Reverse String II (3分)
539 Minimum Time Difference (5分)
536 Construct Binary Tree from String (6分)
527 Word Abbreviation (8分)
539 Minimum Time Difference
Problem:Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list.
Example
Input: [“23:59”,”00:00”]
Output: 1
Note
1 The number of time points in the given list is at least 2 and won’t exceed 20000.
2 The input time is legal and ranges from 00:00 to 23:59.
这是一道中等难度的题,但我觉得破解它并不难,而且实际的AC率也高达50%,当时在思考时,觉得难点主要在于对0点这个点该怎么区分,而恰恰Example中给出了这样的例子,否则还真的不会注意到要区分0点。
所以还是AC了,分享下自己的思考过程吧,我看到时钟之间的差,第一个想到的是每个timepoints分布在了一个圆圈上,无非就是求两个相邻point的最小间隔罢了。但有个问题时,在圆的0时刻上,该怎么判断呢?
首先求points之间的最小间隔,那么points 最好是有序的。那么我就把所有的points依次排序在圆圈上就好了,0时刻有点蛋疼,但我想到的是把圆拉成一条线,呵呵,不就是一个数组么,无非再让最后一个元素和头一个元素再比较一下就好了,就形成了一个环的min diff。So
My first solution(37ms)
public int findMinDifference(List<String> timePoints) { int[] times = new int[timePoints.size()]; int index = 0; for (String num : timePoints){ String[] time = num.split(":"); int minutes = Integer.parseInt(time[0]) * 60 + Integer.parseInt(time[1]); times[index++] = minutes; } Arrays.sort(times); int min = Integer.MAX_VALUE; for (int i = 1; i < times.length; i++){ if(times[i] - times[i-1] < min){ min = times[i] - times[i-1]; } } //最后一个数 if ((times[0] + 24 * 60 )- times[times.length -1] < min){ min = (times[0] + 24 * 60 )- times[times.length -1]; } return min; }
String 转下int数组,方便后期进行diff操作,该算法的时间复杂度为O(nlgn),先对times数组中的头置尾的元素操作,最后比较一次尾和头的元素,但尾头元素,注意要加上一个1440的值,其他没什么。
可优化空间:
1. int型的times数组其实没必要,浪费了空间。
2. 最后一个数的比较和前n-1个数的比较不能写到一块么?
My second solution(44ms)
public int findMinDifference(List<String> timePoints) { int[] times = new int[timePoints.size()]; int index = 0; for (String num : timePoints) { String[] time = num.split(":"); int minutes = Integer.parseInt(time[0]) * 60 + Integer.parseInt(time[1]); times[index++] = minutes; } Arrays.sort(times); int min = Integer.MAX_VALUE; int prev = times[times.length-1]; for (int i = 0; i < times.length; i++){ int diff = (times[i] - prev); if(diff < 0) diff += 1440; min = Math.min(min, diff); prev = times[i]; } return min; }
呵呵,添加了一个prev指针后,实现了最后元素和前n-1个元素的合并,的确蛮巧妙的,算是算法细节吧,真正做题哪顾得上那么多啊。但这里的循环检验着实是一个好的想法,起码构成了一个loop的diff问题吧!注意for循环中的if判断,难道是它影响了运行时间!!!还没优化空间呢,其实有了prev,空间优化的思路已经出来了,直接遍历timePoints解析呗。
My third solution(97ms)
public int findMinDifference(List<String> timePoints) { public int findMinDifference(List<String> timePoints) { Collections.sort(timePoints); int minDiff = Integer.MAX_VALUE; String prev = timePoints.get(timePoints.size() - 1); for (String s : timePoints) { int prevMins = Integer.parseInt(prev.split(":")[0]) * 60 + Integer.parseInt(prev.split(":")[1]); int curMins = Integer.parseInt(s.split(":")[0]) * 60 + Integer.parseInt(s.split(":")[1]); int diff = curMins - prevMins; if (diff < 0) diff += 1440; minDiff = Math.min(minDiff, diff); prev = s; } return minDiff; }
原来,Collections可以直接对timePoints排序啊,真是看了答案才知道的。空间复杂度下降了,但运行时间反而升了。我能想到的就这里了,但在赛题复盘时,发现还有个O(n)解,处于好奇,咱继续研究下。
针对一个排序算法它能形成O(n)解的情况,只能是一个利用有限大小的位置做bucket排序,这让我在解排序问题时,给出了一个新的思考角度,如果是排序,请先思考能否bucket排序,能否bucket排序,能否bucket排序。分钟能出现的位置只能是从[0-1439]中间的任何位置,So
My forth solution(19ms)
public int findMinDifference(List<String> timePoints) { boolean[] timeSeen = new boolean[1440]; for (String s : timePoints) { int mins = Integer.parseInt(s.split(":")[0]) * 60 + Integer.parseInt(s.split(":")[1]); if (timeSeen[mins]) return 0; timeSeen[mins] = true; } int minDiff = Integer.MAX_VALUE,prev = 0; for (int i = 1439; i >= 0; i--){ if(timeSeen[i]){ prev = i; break; } } for (int i = 0; i < 1440; i++) { if(timeSeen[i]){ int diff = i - prev; if(diff <0) diff += 1440; minDiff = Math.min(minDiff, diff); prev = i; } } return minDiff; }
沿用第三种prev的解决方案,我们现在无非就是把它进过一次遍历就放在了指定的位置上,同样的先找出尾部的第一个True元素,与第0个元素比较,剩余元素均与前一个True元素比较,取最小的diff,即为我们的答案。
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