LeetCode 275. H-Index II
275. H-Index II
Description Submission Solutions
- Total Accepted: 42241
- Total Submissions: 125172
- Difficulty: Medium
- Contributors: Admin
Follow up for H-Index: What if the citations
array is sorted in ascending order? Could you optimize your algorithm?
Hint:
- Expected runtime complexity is in O(log n) and the input is sorted.
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【题目分析】
相比上个题目,这次直接给定了一个有序的序列。要求以O(logn)的时间复杂度返回结果。
【思路】
对于一个有序的序列,我们自然想到的就是二分查找算法。如何来做呢?我们画图分析一下:
1. 上图是一个递增的序列,下面一行是此处若满足条件对应的h值。可见h值是降序序列。因此在二分查找的过程中,如果某个h值满足条件(即h值小于它对应的值),我们就到前半部分继续查找;如果h值不满足条件(即h值大于它对应的值),我们就到前半部分继续查找。
2. 对于一些边界情况,如果在最左边查找成功或者失败,则此时left=0,h=len(即len-left);如果在最右边查找失败,left会溢出,left=len,此时h=len-left=0.
【java代码】
1 public class Solution { 2 public int hIndex(int[] citations) { 3 int len = citations.length; 4 5 int left = 0, right = len-1; 6 while(left <= right) { 7 int mid = left+(right-left)/2; 8 if(citations[mid] >= len-mid) right = mid - 1; 9 else left = mid + 1; 10 } 11 return len - left; 12 } 13 }