Title
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：返回最大元素，这个最大元素满足大于等于第二大的二倍，返回最大元素位置，否则返回-1
Solution
自己写的如下
//cpp:
class Solution {
public:
int dominantIndex(vector<int>& nums) {
int maxn = 0, secmax = 0;
int index = -1;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] >= maxn) {
maxn = nums[i];
index = i;
}
}
for (int i = 0; i < nums.size(); i++) {
if (nums[i] < maxn) {
secmax = max(secmax, nums[i]);
}
}
if (maxn >= (2 * secmax))
return index;
else
return -1;
}
};
我的步骤有点多，可以更简单点的
class Solution {
public int dominantIndex(int[] nums) {
int maxIndex = 0;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] > nums[maxIndex])
maxIndex = i;
}
for (int i = 0; i < nums.length; ++i) {
if (maxIndex != i && nums[maxIndex] < 2 * nums[i])
return -1;
}
return maxIndex;
}
}