原题链接:162. Find Peak Element
58 / 58
test cases passed. Runtime: 0 ms Your runtime beats 33.05% of javasubmissions.
58 / 58
test cases passed. Runtime: 52 ms Your runtime beats 94.19% of pythonsubmissions.
【思路-Java、Python】 二分查找
题目中已经说明,最左端和最右端元素均无限小,中间元素比两侧都要大,那么本题中一定存在一个峰元素。所以不管中间有多少波峰,只要找到峰元素,我们只需找到刚刚开始下降而未下降的位置。采用二分查找,查出这样一个位置即可,我们知道二分查找要比较的是 target 元素,本题的 target 元素是 mid 的后一个元素,即 nums[mid] 与 nums[mid+1] 进行比较:
- public class Solution {
- public int findPeakElement(int[] nums) {
- int left = 0, right = nums.length - 1;
- while (left < right) {
- int mid = (left + right) / 2;
- if(nums[mid] < nums[mid + 1]) left = mid + 1;
- else right = mid;
- }
- return left;
- }
- }
- class Solution(object):
- def findPeakElement(self, nums):
- """
- :type nums: List[int]
- :rtype: int
- """
- left, right = 0, len(nums) - 1
- while left < right :
- mid = (left + right) / 2
- if nums[mid] < nums[mid + 1] : left = mid + 1
- else : right = mid
- return left