A peak element is an element that is greater than its neighbors.
Given an input array nums, where nums[i] ≠ nums[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 nums[-1] = nums[n] = -∞.
Example 1:
Input: nums = [1,2,3,1] Output: 2 Explanation: 3 is a peak element and your function should return the index number 2.Example 2:
Input: nums = [1,2,1,3,5,6,4] Output: 1 or 5 Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.Note:
Your solution should be in logarithmic complexity.
要求找到任意一个峰
class Solution { //随机找到一个峰 public int findPeakElement(int[] nums) { int N = nums.length; //在二分查找的过程汇中,以下条件始终成立： //对于nums[left,right], 有nums[left-1]<nums[left] 且 nums[right]>nums[right+1] int left = 0, right = N - 1;//初始时,条件成立 while (right - left > 1) {// int mid = left + (right - left) / 2; //查看nums[mid]和nums[mid+1]的大小关系（根据题意，它们不可能相等） if(nums[mid] < nums[mid + 1]) { left = mid + 1;//nums[mid+1,right]满足条件 } else if(nums[mid] > nums[mid + 1]){ right = mid;//nums[left,mid]满足条件 } } //right==left或者right==left+1 //right==left则直接返回这个元素即可 //right==left+1则需要比较left和right的大小 return (left == N - 1 || nums[left] > nums[left + 1]) ? left : right; } }
