162. Find Peak Element

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.

Approach #1:

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int len = nums.size();
        if (len == 1) return 0;
        int start = 1, end = len-2;
        while (nums[0] == nums[start] && start < len) start++;
        if (nums[0] > nums[start]) return 0;
        while (nums[len-1] == nums[end] && end >= 0) end--;
        if (nums[len-1] > nums[end]) return len-1;
        for (int i = start; i <= end; ++i) {
            if (nums[i] > nums[i-1] && nums[i] > nums[i+1]) return i;
        }
    }
};

Runtime: 8 ms, faster than 14.40% of C++ online submissions for Find Peak Element.

Approach #2: Linear Scan

time: O(n)　　space: O(1)

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int len = nums.size();
        for (int i = 0; i < len-1; ++i) {
            if (nums[i] > nums[i+1])
                return i;
        }
        return len-1;
    }
};

Runtime: 4 ms, faster than 99.33% of C++ online submissions for Find Peak Element.

Approach #3: Recursive Binary Search

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int len = nums.size();
        return search(nums, 0, len-1);
    }
    
    int search(vector<int> nums, int l, int r) {
        if (l == r) return l;
        int m = l + (r - l) / 2;
        if (nums[m] > nums[m+1]) return search(nums, l, m);
        return search(nums, m+1, r);
    }
};

Runtime: 4 ms, faster than 99.33% of C++ online submissions for Find Peak Element.

Approach #4: Iterative Binary Search

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int len = nums.size();
        int l = 0, r = len - 1;
        while (l < r) {
            int m = l + (r - l) / 2;
            if (nums[m] > nums[m+1])
                r = m;
            else
                l = m + 1;
        }
        return l;
    }
    
};

Runtime: 4 ms, faster than 99.33% of C++ online submissions for Find Peak Element.

when i modify the condition with:

        while (l <= r) {
            int m = l + (r - l) / 2;
            if (nums[m] > nums[m+1])
                r = m - 1;
            else
                l = m + 1;
        }

it will report error.

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Analysis:

https://leetcode.com/problems/find-peak-element/