[LeetCode] 162. Find Peak Element 查找峰值元素

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.

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Note:

Your solution should be in logarithmic complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

给一个数组，寻找里面的峰值元素。峰值是比它两边的元素都大。

提示了用log的时间复杂度，所以考虑用二分法Binary Search。

规律一：如果nums[i] > nums[i+1]，则在i之前一定存在峰值元素
规律二：如果nums[i] < nums[i+1]，则在i+1之后一定存在峰值元素

参考：Orange橘子洲头

Java：

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;  
    }  
}  　　

Python：

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 - left) / 2
            if nums[mid] > nums[mid + 1]:
                right = mid
            else:
                left = mid + 1
       
        return left

C++:

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int left = 0, right = nums.size() - 1;
        
        while (left < right) {
            const auto mid = left + (right - left) / 2;
            if (nums[mid] > nums[mid + 1]) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
       
        return left;
    }
};

　　　　

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