C 0 ms O(N) Solution


  • 6
    H
    int hIndex(int* citations, int citationsSize) {
        int i, j = 0, index = 0, *count = malloc(sizeof(int) * (citationsSize + 1));
        memset(count, 0, sizeof(int) * citationsSize);
        for (i = 0; i < citationsSize; ++i) {
            count[citations[i] > citationsSize ? citationsSize : citations[i]]++;
        }
        for (i = 0; i <= citationsSize; ++i) {
            if (citationsSize - j >= i) {
                index = i;
            }
            j += count[i];
        }
        free(count);
        return index;
    }

  • -5
    S

    Hi,

    I came across a situation: given citations = [3, 1, 7, 8, 9]. What will be the answer be?

    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."

    When h=3, the definition will be "A scientist has index 3 if 3 of his/her 5 papers have at least 3 citations each, and the other 2 papers have no more than 3 citations each." Obviously wrong....

    When h=4, the definition will be "A scientist has index 4 if 4 of his/her 5 papers have at least 4 citations each, and the other 1 papers have no more than 4 citations each." Obviously wrong....

    When h=2, the definition will be "A scientist has index 2 if 2 of his/her 5 papers have at least 2 citations each, and the other 3 papers have no more than 3 citations each." Obviously wrong....


  • 2
    K

    line 3 is wrong,it should be
    memset(count, 0, sizeof(int) * (citationsSize+1));


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