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].
分析：找到最大值maxn、它对应的下标idx和次大值sec，如果次大值sec的两倍比maxn大说明不满足条件返回-1，否则返回idx
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class Solution { public: int dominantIndex(vector<int>& nums) { int maxn = INT_MIN, idx = -1, sec = INT_MIN; for (int i = 0; i < nums.size(); i++) { if (nums[i] > maxn) { sec = maxn; maxn = nums[i]; idx = i; } else if(nums[i] > sec){ sec = nums[i]; } } return sec * 2 > maxn ? -1 : idx; } }; |
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