Largest Number Greater Than Twice of Others
Largest Number Greater Than Twice of Others
LWC 64: 747. Largest Number Greater Than Twice of Others
Problem:
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.
Exmaple 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].
思路1： 非排序，如果存在一个数，比其他任何数的两倍还大必然是最大数。时间复杂度为O(n2)O(n^2)
Java版本：
public int dominantIndex(int[] nums) {
int n = nums.length;
for (int i = 0; i < n; ++i) {
boolean ok = true;
for (int j = 0; j < n; ++j) {
if (i != j && nums[i] < nums[j] * 2) {
ok = false;
break;
}
}
if (ok) return i;
}
return -1;
}思路2： 数据量大的情况下，采用排序，最大的数和次大的数比较一波即可，注意下，排序需要带着下标。
Java版本：
class P implements Comparable<P>{
int idx;
int val;
P(int idx, int val){
this.idx = idx;
this.val = val;
}
@Override
public int compareTo(P that) {
return that.val - this.val;
}
}
public int dominantIndex(int[] nums) {
int n = nums.length;
if (n == 1) return -1;
P[] ps = new P[n];
for (int i = 0; i < n; ++i) {
ps[i] = new P(i, nums[i]);
}
Arrays.sort(ps);
if (ps[0].val >= ps[1].val * 2) return ps[0].idx;
return -1;
}Python版本：
def dominantIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
idx = nums.index(max(nums))
nums = sorted(nums)
return idx if nums[-1] >= nums[-2] * 2 else -1