leetcode 747. Largest Number Greater Than Twice of Others 最大元素的Index-蒲公英云

leetcode 747. Largest Number Greater Than Twice of Others 最大元素的Index

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].

本题题意很简单，直接暴力去做即可

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
using namespace std;
class Solution 
{
public:
    int dominantIndex(vector<int>& nums) 
    {
        int largest = nums[0];
        int index = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            if (nums[i] > largest)
            {
                largest = nums[i];
                index = i;
            }
        }
        for (int i = 0; i < nums.size(); i++)
        {
            if (i != index && largest < 2 * nums[i])
                return -1;
        }
        return index;
    }
};