Leetcode#747. Largest Number At Least Twice of Others

最新推荐文章于 2018-10-30 08:46:09 发布

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本文介绍了一种算法，用于在整数数组中找出是否存在一个元素，该元素的值至少是数组中其他所有元素最大值的两倍。若存在，则返回该元素的索引；否则返回-1。文中提供了C++及Python实现代码。

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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].

题意

从数组中找到一个数，此数 >=剩余任意数的二倍，即此数>=剩余最大数的二倍

题解

先找到最大值，判断最大值是否大于次大值的二倍。

C++代码

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        vector<int>numscopy;
        int len = nums.size();
        for(int i=0; i<len; i++){
            numscopy.push_back(nums[i]);
        }
        sort(numscopy.begin(), numscopy.end());

        if(numscopy[len-1] >= numscopy[len-2]*2){
            for(int i=0; i<len; i++){
                if(numscopy[len-1] == nums[i]){
                    return i;
                }
            }
        }
        return -1;

    }
};

python代码

class Solution(object):
    def dominantIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        numscopy = nums[::]
        numscopy.sort()
        n = len(nums)
        if n == 1:
            return 0
        if numscopy[n-1] >= numscopy[n-2]*2:
            return nums.index(numscopy[n-1])
        return -1