[Leetcode] 747. Largest Number At Least Twice of Others 解题报告_in a given integer array nums, there is always exa-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79536910

本文介绍了一种算法，用于在整数数组中找到是否有一个元素至少是其他所有元素值两倍的方法，并返回该元素的索引。若不存在则返回-1。通过一次遍历记录最大值与次大值，实现高效查找。

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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.

Example 2:

Input: nums = [1, 2, 3, 4]
Output: -1
Explanation: 4 isn't at least as big as twice the value of 3, so we return -1.

Note:

nums will have a length in the range [1, 50].
Every nums[i] will be an integer in the range [0, 99].

思路：

如果采用两遍扫描的话，我们第一遍就找出数组中的最大值，然后第二遍判断其余数字是否小于等于最大值的1/2。但是实际上我们可以一遍扫描解决问题：记录下数组中的最大值和次大值，最后判断一下最大值是否大于等于次大值的二倍即可。

代码：

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        if (nums.size() == 0) {
            return -1;
        }
        else if (nums.size() == 1) {
            return 0;
        }
        int largest_index = nums[0] > nums[1] ? 0 : 1;
        int larger_index = 1 - largest_index;
        for (int i = 2; i < nums.size(); ++i) {
            if (nums[i] > nums[largest_index]) {
                larger_index = largest_index;
                largest_index = i;
            }
            else if (nums[i] > nums[larger_index]) {
                larger_index = i;
            }
        }
        return nums[larger_index] * 2 <= nums[largest_index] ? largest_index : -1;
    }
};