题目

In a given integer array nums, there is always exactly one largest element.

Find whether the largest element in the array is at least twice as much as every other number in the array.

If it is, return the index of the largest element, otherwise return -1.

Example 1:

Input: nums = [3, 6, 1, 0]
Output: 1
Explanation: 6 is the largest integer, and for every other number in the array x,
6 is more than twice as big as x. The index of value 6 is 1, so we return 1.

我的想法

用max存最大值序列，lastmax存第二大序列。但将max和lastmax初始化为0，如果num[0]的值就是最大值，lastmax和max都是最大序列号，lastmax无法更新

//记录错误做法
class Solution {
    public int dominantIndex(int[] nums) {
        if(nums.length == 1) return 0;
        int max = 0, lastMax = 0;
        for(int i = 0; i < nums.length; i++){
            if(nums[i] > nums[lastMax]) lastMax = i;
            if(nums[i] > nums[max]){
                lastMax = max;
                max = i;                
            }
        }
        if(nums[max] >= 2*nums[lastMax]) return max;
        else return -1;
    }
}

根据答案提示修改：lastmax不存序号，直接保存值的大小，因为max是在序号0，循环直接从i=1开始

//修改正确
class Solution {
    public int dominantIndex(int[] nums) {
        if(nums.length == 1) return 0;
        int max = 0, lastMax = -1;
        for(int i = 1; i < nums.length; i++){
            if(nums[i] > lastMax) lastMax = nums[i];
            if(nums[i] > nums[max]){
                lastMax = nums[max];
                max = i;                
            }
        }
        if(nums[max] >= 2*lastMax) return max;
        else return -1;
    }
}

解答

记录最大和第二大的值，以及最大值的序号

class Solution {
    public int dominantIndex(int[] nums) {
        int max = -1, index = -1, second = -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > max) {
                second = max;
                max = nums[i];
                index = i;
            } else if (nums[i] > second)
                second = nums[i];
        }
        return second * 2 <= max ? index : -1;
    }
}