747. Largest Number At Least Twice of Others

最新推荐文章于 2025-12-14 15:24:42 发布

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原文链接：http://www.cnblogs.com/Zzz-y/p/8127789.html

本文介绍了一种算法，用于寻找数组中是否有一个元素至少是其他所有元素值两倍的情况，并返回该元素的索引。通过一次遍历找到最大值及其次大值，以此判断条件是否成立。

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:

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.

找数组中的最大值，并判断是否至少是其他数的两倍。是的话返回索引，不是的话返回-1.

解决：遍历数组，找最大值m1和次大值m2，记录最大值坐标。只要 m1>=2*m2 ，那么就必然符合条件，返回坐标位置。不然就返回-1.

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        int m1 = INT_MIN;
        int m2 = INT_MIN;
        int key = -1;
        for (int i=0; i<nums.size(); ++i) {
            if (nums[i]>m1) {
                m2 = m1;
                m1 = nums[i];
                key = i;
            }
            else if (nums[i]>m2)
                m2 = nums[i];
        }
        return m1>=2*m2? key:-1;
    }
};