# h引用（数组中有h个数大于等于h，其余小于）H-Index

2017/12/19 10:49

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."

For example, given `citations = [3, 0, 6, 1, 5]`, which means the researcher has `5` papers in total and each of them had received `3, 0, 6, 1, 5` citations respectively. Since the researcher has `3` papers with at least `3` citations each and the remaining two with no more than `3` citations each, his h-index is `3`.

Note: If there are several possible values for `h`, the maximum one is taken as the h-index.

【题意】有h篇论文的引用数目大于等于h，其余引用小于h。

① 排序法。先将数组排序，我们就可以知道对于某个引用数，有多少文献的引用数大于这个数。对于引用数`citations[i]`，大于该引用数文献的数量是`citations.length - i`，而当前的H-Index则是`Math.min(citations[i], citations.length - i)`，我们将这个当前的H指数和全局最大的H指数来比较，得到最大H指数。

1)

class Solution { //4ms
public int hIndex(int[] citations) {
Arrays.sort(citations);
int h = 0;//全局最大的H指数
for (int i = 0;i < citations.length;i ++){
int curH = Math.min(citations[i],citations.length - i);//得到当前的H指数
if (curH > h){
h = curH;
}
}
return h;
}
}

2)

class Solution{//2ms
public int hIndex(int[] citations) {
Arrays.sort(citations);
for (int i = 0; i < citations.length; i ++) {
if (citations[i] >= citations.length - i) {
return citations.length - i;
}
}
return 0;
}
}

② 数组映射法。我们额外使用一个大小为N+1的数组dp。dp[i]表示有多少文章被引用了i次，这里如果一篇文章引用大于N次，我们就将其当为N次，因为H指数不会超过文章的总数。

1)

class Solution { //1ms
public int hIndex(int[] citations) {
int len = citations.length;
int[] dp = new int[len + 1];
for (int i = 0;i < len;i ++){//统计各个引用次数对应多少篇文章
int index = citations[i] <= len ? citations[i] : len;
dp[index] += 1;
}
int count = 0;
for (int i = len;i > 0;i --){
count+= dp[i];
if (count >= i){
return i;
}
}
return 0;
}
}

2)

class Solution { //1ms
public int hIndex(int[] citations) {
int[] dp = new int[citations.length + 1];
for (int n : citations){
if (n >= citations.length){
dp[citations.length] ++;
}else{
dp[n] ++;
}
}
int count = 0;
for (int i = citations.length;i > 0;i --){
count += dp[i];
if (count >= i){
return i;
}
}
return 0;
}
}

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