[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.

給一個數組，尋找里面的峰值元素。峰值是比它兩邊的元素都大。

提示了用log的時間復雜度，所以考慮用二分法Binary Search。

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

參考：Orange橘子洲頭

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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