如何在数组中找到第二大数字，但返回该值出现的最后一个索引？

Question

My actual problem is to find the number in a data set with the 2nd highest frequency. I'm initializing an array that is the size of the largest number in the data set and then incrementing the corresponding indexes in the array each time the number appears in the data set. If more than two indexes shares the 2nd highest frequency then I need to return the larger index. My code returns the correct answer in some cases, but not all and I am having trouble finding the error in my logic.

int secondFreq(int data[], int maxNum){
    int highest = 0;
    int secondHighest = 0;
    int x;

    for(x = 0; x < maxNum; x++){
        if(data[x] > data[highest]){
            secondHighest = highest;
            highest = x;
        }
        else if (data[x] > data[secondHighest]){
            secondHighest = x;
        }
        else if (data[x] == data[secondHighest]){
            secondHighest = x;
        }
    }

    return secondHighest + 1;
}

Here is an example of an array that yields the wrong answer. The left number is the index and the right number is the value stored at that index. My function returns 12 (11th index+1), but it should be returning 5 (4th index + 1).

    0 - 2
    1 - 2
    2 - 5
    3 - 2
    4 - 5
    5 - 4
    6 - 2
    7 - 2
    8 - 6
    9 - 4
    10 - 3
    11 - 6
    12 - 2
    13 - 2
    14 - 3

Answer 1

In your example, the second '6'(index 11) will go to the 'else if (data[x] > data[secondHighest])' and update the 'secondHighest' to be the same as 'highest'. This logic cannot deal with the condition where there are multiple highest value. To fix, you can put a 'else if(data[x] == data[highest])' before other 'else if'.

However, you will get another incorrect anwser if your data set is as below:

0 - 2
1 - 2
2 - 5

I will change the code to be more clear and readable as below:

int secondFreq(int data[], int maxNum){
    int secondLargestValue = findsecondLargestValue(data);
    return findIndexofValue(data, secondLargestValue) + 1;
}

int findsecondLargestValue(int data) {
    ...
}

int findIndexofValue(int data, int value) {
    ...
}

Answer 2

Were it not for the efficiency consideration of being 2 pass, (so this is not going to be the best answer), I would want to say something like:

int secondFreq(int data[], int maxNum)
{
    double          highestVal = data[0];   // start with the first
                        // entry
    int             secondHighest = -1; // will send back 0 if everything
                    // ties for first
    int             x;      // for loop index

    for (x = 1; x < maxNum; x++) {
    if (data[x] > highestVal) {
        highestVal = data[x];
    }
    };
    for (x = 1; x < maxNum; x++) {
    if (data[x] != highestVal)
        &&((secondHighest == -1) || (data[x] >= data[secondHighest])) {
        secondHighest = x;
        }
    }


    return secondHighest + 1;
}

It is harder to mess up, though it does look through twice.

Answer 3

Your code has two problems:

1. You will need to set highest = x; in the case of data[x] == data[highest] .
2. secondHighest should be initially pending.

The following is an example modification:

int secondFreq(int data[], int maxNum){
    int highest = 0;
    int secondHighest, secondFlag = 0;//secondFlag : Whether secondHighest has been determined, 0 : undetermined, 1(Not 0) : determined
    int x;

    for(x = 1; x < maxNum; x++){
        if(data[x] > data[highest]){
            secondHighest = highest;
            highest = x;
            secondFlag = 1;
        }
        else if (data[x] == data[highest]){
            highest = x;
        }
        else if (secondFlag == 0 || data[x] >= data[secondHighest]){
            secondHighest = x;
            secondFlag = 1;
        }
    }

    if(secondFlag)
        return secondHighest + 1;
    else
        return 0;//All elements same
}