747. Largest Number At Least Twice of Others

最新推荐文章于 2020-10-03 10:33:35 发布

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最新推荐文章于 2020-10-03 10:33:35 发布

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分类专栏： leetcode 算法 Leetcode 文章标签： leedcode

本文链接：https://blog.youkuaiyun.com/qq_36718317/article/details/79252502

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本文介绍了一种算法，用于在整数数组中找到最大元素，并判断该元素是否至少是其他每个元素值的两倍。如果是，则返回该元素的索引；否则返回-1。文章通过实例展示了如何实现这一功能。

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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].

题目大意：找到vector中最大的数，若它比其他的数都至少大一倍就返回它的位置，否则返回-1。

思路：

可以用一个map，first存放实际值，second存放索引。若最大的数不小于次大的数的两倍，那么就返回它的索引，否则返回-1.注意如果只有一个数就返回0

代码：

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        map<int,int>m;
        int n=nums.size();
        for(int i=0;i<n;i++)
            m.insert(pair<int,int>(nums[i],i));
        int t=m.rbegin()->first;
        int h=m.rbegin()->second;
        if(n<2)
            return 0;
        for(map<int,int>::reverse_iterator rit=m.rbegin();rit!=m.rend();rit++)
        {
            if(rit==m.rbegin())
                rit++;
            if(t<2*((rit)->first))
                return -1;
            else
                return h;
        }
    }
};