A peak element is an element that is greater than its neighbors.
Given an input array where num[i] ≠ num[i+1], find a peak element and return
its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that num[-1] = num[n] = -∞.
For example, in array [1, 2, 3, 1], 3 is a peak element and your function
should return the index number 2.
找数组中的任意一个峰值(返回其下标):比旁边两个元素都要大即可。可以假设num[-1] = num[n] = -∞.
这题的意思应该就是峰值一定存在了。不会出现[1,1,1]或者[0,1,1,0]这样的情况。
遍历一遍数组当然可以,但是用二分查找的话时间降到O(logN):
class Solution {
public:
int findPeakElement(const vector<int> &num)
{
int low = 0;
int high = num.size()-1;
while(low < high)
{
int mid1 = (low+high)/2;
int mid2 = mid1+1;
if(num[mid1] < num[mid2])
low = mid2;
else
high = mid1;
}
return low;
}
};
本文介绍了一种使用二分查找算法在数组中找到一个峰值元素的方法,峰值元素定义为大于其邻居的元素。通过二分查找可以在O(logN)的时间复杂度内找到符合条件的元素。
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