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本文链接:https://blog.csdn.net/shiliang97/article/details/103180783
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].
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/largest-number-at-least-twice-of-others 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution {
public:
int dominantIndex(vector<int>& nums) {
int maxid=0;
int max=nums[0];
int nextid=0;
int next=0;
for(int i=1;i<nums.size();i++){
if(nums[i]>max){
next=max;
nextid=maxid;
maxid=i;
max=nums[i];
}else if(nums[i]>next){
nextid=i;
next=nums[i];
}
}if(max>=2*next){
return maxid;
}return -1;
}
};