Leetcode OJ 274. H-index

Given an array of citations (each citation was a non-negative integer) of a researcher, write a function to compute the RES Earcher ' s h-index.

According to the definition of H-index on Wikipedia: ' A scientist has index H if h of his/her N Papers has at least H citations each, and the other N? h Papers has no more than H citation S each. "

For example, given citations = [3, 0, 6, 1, 5] , which means the researcher have papers in total and each of 5 them had received 3, 0, 6, 1, 5 Citatio NS respectively. Since the researcher have papers with at least citations each and the remaining both with 3 3 no more than 3 CIT Ations each, he h-index is 3 .

Note:if there was several possible values h for, the maximum one is taken as the h-index.

is a quick row, but test instructions a bit vague, at least h citations does not necessarily mean that there must be equal to H citations, and the remaining n-h is no more than, that is, can exist equal to the situation.

voidSwap (intA[],intAintBB) {    intTMP =A[a]; A[a]=A[b]; A[B]=tmp;}intPartition (int* Citations,intLeftintRight ) {    intPivot =Citations[right]; intI, j =0;  for(i =0; I < right; i++){        if(Citations[i] >=pivot)            {Swap (citations, I, j); J++;        }} Swap (citations, J, right); returnJ;}voidQuick_sort (int* Citations,intLeftintRight ) {    intpivot_position; if(Left >=Right ) {        return; } pivot_position=Partition (citations, left, right); Quick_sort (citations, left, Pivot_position-1); Quick_sort (citations, pivot_position+1, right);}intHindex (int* Citations,intcitationssize) {    inti; Quick_sort (citations,0, Citationssize-1);  for(i = citationssize-1; I >=0; i--){        if(Citations[i] >= i +1){                returni +1; }    }    return 0;}

Leetcode OJ 274. H-index