leetcode [274] - H-Index
Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.
According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each."
Example:
Input:citations = [3,0,6,1,5]
Output: 3 Explanation:[3,0,6,1,5]
means the researcher has5
papers in total and each of them had received3, 0, 6, 1, 5
citations respectively. Since the researcher has3
papers with at least3
citations each and the remaining two with no more than3
citations each, her h-index is3
.
Note: If there are several possible values for h, the maximum one is taken as the h-index.
题目大意:
给定一位研究者论文被引用次数的数组,输出研究者的 h 指数。h指数指他的 (N 篇论文中)至多有 h 篇论文分别被引用了至少 h 次。
理 解:
首先对数组从大到小排序,使用二分法进行比较。
若 当前引用篇数 <= 当前最小引用数量,则向右继续找。保存当前指数=当前引用篇数,更新最大h指数。
若 当前引用篇数 > 当前最小引用数量,则向左找。
代 码 C++:
class Solution { public: int hIndex(vector<int>& citations) { int n = citations.size(); if(n<=0) return 0; if(n==1 && citations[0]>=1) { return 1; } int i,j,temp; for(i=0;i<n;++i){ for(j=i;j<n;++j){ if(citations[i]<citations[j]){ temp = citations[i]; citations[i] = citations[j]; citations[j] = temp; } } } if(n<=citations[n-1])return n; int left=0,right=n-1,mid; int h = 0; while(left<right){ mid = (left+right)/2; if(mid+1<=citations[mid]){ if(mid+1>h) h = mid+1; left = mid+1; }else{ right = mid; } } return h; } };