Find Peak Element 解答

Question

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.

Note:

Your solution should be in logarithmic complexity.

Solution

The assumption is a very good hint. It assures that there must be a peak in input array.

We can solve this problem by Binary Search.

Four situations to consider:

1. nums[mid] > nums[mid - 1] && nums[mid] > nums[mid + 1]

=> mid is a peak

2. nums[mid] > nums[mid - 1] && nums[mid] < nums[mid + 1]

=> There must exists a peak in right side

3. nums[mid] < nums[mid - 1] && nums[mid] > nums[mid + 1]

=> There must exists a peak in left side

4. nums[mid] < nums[mid - 1] && nums[mid] < nums[mid + 1]

=> Either in right or left side, there must exists a peak.

 

 1 public class Solution {
 2     public int findPeakElement(int[] nums) {
 3         // Binary Search to find peak
 4         int start = 0, end = nums.length - 1, mid = 0, prev = 0, next = 0;
 5         while (start + 1 < end)  {
 6             mid = (end - start) / 2 + start;
 7             prev = mid - 1;
 8             next = mid + 1;
 9             if (nums[mid] > nums[prev] && nums[mid] > nums[next])
10                 return mid;
11             if (nums[mid] > nums[prev] && nums[mid] < nums[next]) {
12                 start = mid;
13                 continue;
14             }
15             if (nums[mid] < nums[prev] && nums[mid] > nums[next]) {
16                 end = mid;
17                 continue;
18             }
19             start = mid;
20         }
21         if (nums[start] > nums[end])
22             return start;
23         return end;
24     }
25 }

 

转载于:https://www.cnblogs.com/ireneyanglan/p/4889395.html