[LeetCode ] H-Index
阿新 • • 發佈:2018-12-22
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 leasth 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 has 5 papers in total and each of them had received 3, 0, 6, 1, 5citations respectively. Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than3 citations each, her h-index is 3.
Note: If there are several possible values for h, the maximum one is taken as the h-index.
題意:給出一個n個元素的陣列,找出一個h,這個陣列有h個數大於等於h,剩下h個數小於等於h,輸出最大的h。
思路:
解法一:再定義一個數組,記錄每個元素出現了多少次,再對這個陣列求字尾和,然後從後向前遍歷,如果字尾和大於等於i,停止遍歷。時間複雜度O(N)。
C程式碼:
int hIndex(int* citations, int citationsSize) { int ans = 0,i; int pre[1005]; memset(pre,0,sizeof(pre)); for(i = 0; i < citationsSize; i++) { pre[citations[i]]++; } for(i = 1000; i >= 0; i--) { pre[i] = pre[i + 1] + pre[i]; if(pre[i] >= i) { ans = i; break; } } return ans; }
解法二:先對陣列排序,從後向前遍歷,如果此時列舉到的數大於等於已經列舉的數的個數,停止遍歷。時間複雜度O(NlogN)。
Java程式碼:
public int hIndex(int[] citations) {
int ans = 0;
Arrays.sort(citations);
for(int i = citations.length; i >= 0; i--) {
if(citations[i] >= citations.length - i) {
ans = citations.length - i;
break;
}
}
return ans;
}