Much Larger Numbers

Input : an array of integers
Output: *index* of the "Much Larger" number, or -1 if there is no such number.

A number is "Much Larger" if it is more than twice every other value in the array.


1,6,19,9: should return "2", the index of 19
1,11,6,9: should return "-1", because there is no "Much Larger" number.
1,19,19,9: should return "-1", because there is no single value that is "much larger".

思路

• 一个是过一遍找最大值再比较
• 一个是one for loop就解决, 找second max

public class MuchLargetNum {

    public int muchLarger(int[] input) {
        if (input == null || input.length == 0) {
            return -1;
        }

        int maxNumber = input[0];
        //int countMax = 1;
        int indexOfMax = 0;
        int n = input.length;

        for (int i = 1; i< n; i++) {
            if (input[i] == maxNumber) {
                //countMax++;
            } else if (input[i] >= maxNumber) {
                maxNumber =input[i];
                indexOfMax = i;
                // countMax = 1;
            }
        }

    //if (countMax > 1) {
    //    return -1;
    //}

        for (int i = 0; i < n; i++) {
            if (indexOfMax == i) {
            continue;
            } else {
                if ((maxNumber / 2.0) <= (input[i])){
                    return -1;
                }
            }
        }
            return indexOfMax;
    }
}